Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
Physics help please
Answer:
i think the answer is 0.001m³
two identical eggs are dropped from the same height. The first eggs lands on a dish and breaks, while the second lands on a pillow and does not break. Which quantities are the same in both situations
Answer:
The height is the same
Explanation:
Because they were at the same height but they fell at different velocities
Define relative density.
Relative density is the ratio of the density of a substance to the density of a given material.
The energy truck travelling at 10 km/h has kinetic energy. How much kinetic energy does it have when it is loaded so its mass is twice and its speed is increased to twice?
Explanation:
The initial kinetic energy [tex]KE_0[/tex] is
[tex]KE_0 = \frac{1}{2}m_0v_0^2[/tex]
When its mass and velocity are doubled, its new kinetic energy KE is
[tex]KE = \frac{1}{2}(2m_0)(2v_0)^2 = \frac{1}{2}(2m_0)(4v_0^2)[/tex]
[tex]\:\:\:\:\:\:\:= 8 \left(\frac{1}{2}m_0v_0^2 \right)= 8KE_0[/tex]
Therefore the kinetic energy will increase by a factor of 8.
. Set the applied force to Force necessary to Keep the box Moving without accelerating. Restart the animation. Just before the box hits the wall, stop the animation. What can you tell me about relative magnitudes of the frictional force and the applied force
Answer:
elative magnitude of the two forces is the same and they are applied in a constant direction.
Explanation:
Newton's second law states that the sum of the forces is equal to the mass times the acceleration
∑ F = m a
in this case there are two forces on the x axis
F_applied - fr = 0
since they indicate that the velocity is constant, consequently
F_applied = fr
the relative magnitude of the two forces is the same and they are applied in a constant direction.
a girl is moving with a uniform velocity of 1.5 m/s then mathematically find her acceleration
Answer:
0
Explanation:
a = dv/dt
if v is constant than the slope of the v graph will be 0, so dv/dt is 0
a= 0
You are driving in such a way that the car is accelerating at a constant rate in the positive direction. When you pass the first sign, you are traveling at 4 m/s. When you pass the second sign 50 m down the road, you note that the seconds indicator of your clock reads 45 seconds. You also note that your velocity is now 9 m/s.
Required:
a. What is your acceleration?
b. What was the clock’s seconds indicator reading when you passed the first sign?
Answer:
Explanation:
a)
v² = u² + 2 a s
v = 9 m/s
u = 4 m/s
s = 50 m
9² = 4² + 2 x a x 50
a = 0.65 m /s²
Acceleration is 0.65 m /s²
b )
time elapsed before velocity changed from 4 m/s to 9 m/s with acceleration of .65 m /s ²
(v - u ) / t = a
(v - u ) / a = t
(9 - 4 ) / .65 = t
t = 7.7
time when passing the first sign will be 7.7 s earlier .
Reading of time indicator = 45 - 7.7
= 37.3 seconds.
Answer:
(a) 0.45 m/s^2
(b) 33.9 s
Explanation:
initial velocity, u = 4 m/s
final velocity, v = 9 m/s
distance, s = 50 m
(a) Let the acceleration is a.
Use third equation of motion
[tex]v^2 = u^2 + 2 as \\\\9^2 = 4^2 + 2\times a\times 50\\\\a = 0.45 m/s^2[/tex]
(b) Let the time is t.
Use first equation of motion
v = u + at
9 = 4 + 0.45 x t
t = 11.1 s
So, the initial time, t' = 45 - 11.1 = 33.9 s
Explore the Prisms screen to see how your understanding of refraction applies when light travels through a medium like glass. Give specific examples and images from the simulation to explain how your understanding applies
Explanation:
https://tse2.mm.bing.net/th?id=OGC.b52c959ac810db1177599a161631c917&pid=Api&rurl=https%3a%2f%2fupload.wikimedia.org%2fwikipedia%2fcommons%2fthumb%2ff%2ff5%2fLight_dispersion_conceptual_waves.gif%2f266px-Light_dispersion_conceptual_waves.gif&ehk=TdcWPzr5xGP8xUOSOqZXauGOS1jHDMu7WnxPzkl7esw%3d
4. Paper is solid in packets labelled 80 g/m2. This means that a sheet of paper of area
10 000cm? has a mass of 80 g. The thickness of each sheet is 0.11mm. What is the
density of the paper?
A 0.073 g/cm?
B 0.088 g/cm
C 0.73 g/cm3
D 0.88 g/cm
B
с
Answer:
Option C. 0.73 g/cm³
Explanation:
From the question given above, the following data were obtained:
Mass = 80 g
Area (A) = 10000 cm²
Thickness = 0.11 mm
Density =?
Next, we shall convert 0.11 mm to cm. This can be obtained as follow:
10 mm = 1 cm
Therefore,
0.11 mm = 0.11 mm × 1 cm / 10 mm
0.11 mm = 0.011 cm
Thus, 0.11 mm is equivalent to 0.011 cm.
Next, we shall determine the volume of the paper. This can be obtained as follow:
Area (A) = 10000 cm²
Thickness = 0.011 cm
Volume =?
Volume = Area × Thickness
Volume = 10000 × 0.011
Volume = 110 cm³
Finally, we shall determine the density of the paper. This can be obtained as follow:
Mass = 80 g
Volume = 110 cm³
Density =?
Density = mass / volume
Density = 80 / 110
Density = 0.73 g/cm³
Therefore the density of the paper is 0.73 g/cm³
A train moving with a uniform speed covers a distance of 120 m in 2 s. Calculate
(i) The speed of the train
(ii) The time it will taketo cover 240 m.
Answer:
(I)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 120 = (u \times 2) + \frac{1}{2} \times 0 \times {2}^{2} \\ 120 = 2u \\ { \tt{speed = 60 \: {ms}^{ - 1} }}[/tex]
(ii)
[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 240 = (60t) \\ { \tt{time = 4 \: seconds}}[/tex]
If a 1.3 kg mass stretches a spring 4 cm, how much will a 5.8 kg mass stretch the
spring? Show MATH, answer and unit.
Answer:
17.8cm
Explanation:
1.3kg --> 4cm
1kg --> 3, 1/13cm
5.8kg --> 18.8cm
If an object of a constant mass experiences a constant net force, it will have a constant what?
Explanation:
hope it helps !!!!!!!!!!!!!
If an object of a constant mass experiences a constant net force, it will have a constant acceleration.
What is force?The definition of force in physics is: The push or pull on a massed object changes its velocity. An external force is an agent that has the power to alter the resting or moving condition of a body. It has a direction and a magnitude.
The application of force is the location at which force is applied, and the direction in which the force is applied is known as the direction of the force. A spring balance can be used to calculate the Force. The Newton is the SI unit of force.
According to Newton's second law of motion:
Applied force = mass × acceleration.
Hence, if an object of a constant mass experiences a constant net force, it will have a constant acceleration.
Learn more about force here:
https://brainly.com/question/26115859
#SPJ6
The voltage across a membrane forming a cell wall is 74.0 mV and the membrane is 9.20 nm thick. What is the electric field strength in volts per meter
Answer:
7.60× 10^6 V/m
Explanation:
electric field strength can be determined as ratio of potential drop and distance, I.e
E=V/d
Where E= electric field
V= potential drop= 74.0 mV= 0.07 V
d= distance= 9.20 nm = 9.2×10^-9 m
Substitute the values
E= 0.07/ 9.2×10^-9
= 7.60× 10^6 V/m
After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at 19.0 with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits
Answer:
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Explanation:
Given :
The first dark fringe is for m = 0
[tex]$\theta_1 = \pm 19^\circ$[/tex]
Now we know for a double slit experiments , the position of the dark fringes is give by :
[tex]$d \sin \theta=\left(m+\frac{1}{2}\right) \lambda$[/tex]
The ratio of distance between the two slits, d to the light's wavelength that illuminates the slits, λ :
[tex]$d \sin \theta=\left(\frac{1}{2}\right) \lambda$[/tex] (since, m = 0)
[tex]$d \sin \theta=\frac{\lambda}{2}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin \theta}$[/tex]
[tex]$\frac{d}{\lambda} = \frac{1}{2 \sin 19^\circ}$[/tex]
[tex]$\frac{d}{\lambda} = 1.54$[/tex]
Therefore, the ratio is [tex]$\frac{1}{1.54}$[/tex] or 1 : 1.54
What are stepdown transformers used for
Answer:
Step down transformers are used in power adaptors and rectifiers to efficiently decrease the voltage. They are also used in electronic SMPS.
Explanation:
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In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a 120-V receptacle on a wall. The picture tube of the television set uses 76 W, and there is 5.5 mA of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio Ns/Np of the transformer.
Ns/Np = ______.
Answer:
c) N_s / N_p = 115.15
Explanation:
Let's look for the voltage in the secondary, they do not indicate the power dissipated
P = V_s i
V_s = P / i
V_s = 76 / 5.5 10⁻³
V_s = 13.818 10³ V
the relationship between the primary and secondary of a transformer is
[tex]\frac{V_p}{N_p} = \frac{V_s}{N_s}[/tex]
[tex]\frac{N_s}{N_p} = \frac{V_s}{V_p}[/tex]
Ns / Np = 13,818 10³ /120
N_s / N_p = 115.15
ASK YOUR TEACHER A 2.0-kg mass swings at the end of a light string with the length of 3.0 m. Its speed at the lowest point on its circular path is 6.0 m/s. What is its kinetic energy at an instant when the string makes an angle of 50 degree with the vertical
Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
A 150.0-kg crate rests in the bed of a truck that slows from 50.0 km/h to a stop in 12.0 s. The coefficient of static friction between the crate and the truck bed is 0.645. What is the minimum stopping time for the truck in order to prevent the crate from sliding?
By Newton's second law,
• the net force acting vertically on the crate is 0, and
∑ F = n - mg = 0 ==> n = mg = 1470 N
where n is the magnitude of the normal force; and
• the net force acting in the horizontal direction on the crate is also 0, with
∑ F = f - b = 0 ==> b = f = µn = 0.645 (1470 N) = 948.15 N
where b is the magnitude of the braking force, f is (the maximum) static friction, and µ is the coefficient of static friction. This is to say that static friction has a maximum magnitude of 948.15 N. If the brakes apply a larger force than this, then the crate will begin to slide.
Note that we are taking the direction of the truck's motion as it slows down to be the positive horizontal direction. The brakes apply a force in the negative direction to slow down the truck-crate system, and static friction keeps the crate from sliding off the truck bed so that the frictional force points in the positive direction.
Let a be the acceleration felt by the crate due to either the brakes or friction. Use Newton's second law again to solve for a :
f = ma ==> a = (948.15 N) / (150.0 kg) = 6.321 m/s²
With this acceleration, the truck will come to a stop after time t such that
0 = 50.0 km/h - (6.321 m/s²) t ==> t ≈ (13.9 m/s) / (6.321 m/s²) ≈ 2.197 s
and this is the smallest stopping time possible.
A 64-ka base runner begins his slide into second base when he is moving at a speed of 3.2 m/s. The coefficient of friction between his clothes and Earth is 0.70. He slides so that his speed is zero just as he reaches the base.
Required:
a. How much mechanical energy is tout due to friction acting on the runner?
b, How far does he slide?
Answer:
Explanation:
From the given information:
mass = 64 kg
speed = 3.2 m/s
coefficient of friction [tex]\mu =[/tex] 0.70
The mechanical energy touted relates to the loss of energy in the system as a result of friction and this can be computed as:
[tex]W = \Delta K.E[/tex]
[tex]\implies \dfrac{1}{2}m(v^2 -u^2)[/tex]
[tex]= \dfrac{1}{2}(64.0 \kg) (0 - (3.2 \ m/s^2))[/tex]
Thus, the mechanical energy touted = 327.68 J
According to the formula used in calculating the frictional force
[tex]F_r = \mu mg[/tex]
= 0.70 × 64 kg× 9.8 m/s²
= 439.04 N
The distance covered now can be determined as follows:
d = W/F
d = 327.68 J/ 439.04 N
d = 0.746 m
An object is moving from north to south what is the direction of the force of friction of the object
Answer:
North
Explanation:
Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.
Answer:
South To North
Explanation:
Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north
A chimpanzee sitting against his favorite tree gets up and walks 51 m due east and 39 m due south to reach a termite mound, where he eats lunch. (a) What is the shortest distance between the tree and the termite mound
Answer:
64.20m
Explanation:
As we can see from the image I have attached below, the route that the chipanzee makes forms a right triangle. In this case, the shortest distance is represented by x in the image, which is the hypotenuse. To find this value we use the Pythagorean theorem which is the following.
[tex]a^{2} +b^{2} = c^{2}[/tex]
where a and b are the length of the two sides and c is the length of the hypotenuse (x). Therefore, we can plug in the values of the image and solve for x
[tex]51^{2} +39^{2} =x^{2}[/tex]
2,601 + 1,521 = [tex]x^{2}[/tex]
4,122 = [tex]x^{2}[/tex] ... square root both sides
64.20 = x
Finally, we see that the shortest distance is 64.20m
a microwave operates at a frequency of 2400 MHZ. the height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. assume that microwave energy is generated uniformly on the uipper surface. What is the power output of the oven
Complete question is;
A microwave oven operates at a frequency of 2400 MHz. The height of the oven cavity is 25 cm and the base measures 30 cm by 30 cm. Assume that microwave energy is generated uniformly on the upper surface of the cavity and propagates directly
downward toward the base. The base is lined with a material that completely absorbs microwave energy. The total microwave energy content of the cavity is 0.50 mJ.
Answer:
Power ≈ 600,000 W
Explanation:
We are given;
Frequency; f = 2400 Hz
height of the oven cavity; h = 25 cm = 0.25 m
base area; A = 30 cm by 30 cm = 0.3m × 0.3m = 0.09 m²
total microwave energy content of the cavity; E = 0.50 mJ = 0.5 × 10^(-3) J
We want to find the power output and we know that formula for power is;
P = workdone/time taken
Formula for time here is;
t = h/c
Where c is speed of light = 3 × 10^(8) m/s
Thus;
t = 0.25/(3 × 10^(8))
t = 8.333 × 10^(-10) s
Thus;
Power = (0.5 × 10^(-3))/(8.333 × 10^(-10))
Power ≈ 600,000 W
Given a 64.0 V battery and 30.0 Ω and 88.0 Ω resistors, find the current (in A) and power (in W) for each when connected in series. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W (b) Repeat when the resistances are in parallel. I30.0 Ω = A P30.0 Ω = W I88.0 Ω = A P88.0 Ω = W
Answer:
a. i. 0.542 A ii. 8.813 W iii. 0.542 A iv. 25.85 W
b. i. 2.13 A ii. 136.53 W iii. 0.727 A iv. 46.55 W
Explanation:
a. Find the current (in A) and power (in W) for each when connected in series.
Since the resistors are connected in series, their combined resistance is R = R₁ + R₂ where R₁ = 30.0 Ω and R₂ = 88.0 Ω.
So, substituting the values of the variables into the equation, we have
R = R₁ + R₂
R = 30.0 Ω + 88.0 Ω
R = 118.0 Ω
Since from Ohm's law, V = IR where V = voltage across circuit = battery voltage = 64.0 V, I = current in circuit and R = total resistance of circuit = 118.0 Ω
So, I = V/R = 64.0V/118.0 Ω = 0.542 A
Since the resistors are in series, the same current flows through them
i. Current in 30.0 Ω
Current in 30.0 Ω is I = 0.542 A since the resistors are in series.
ii Power in the 30.0 Ω
The power in the 30.0 Ω is P₁ = I²R₁ where I = current = 0.542 A and R₁ = resistance = 30.0 Ω
So, P₁ = I²R₁
= (0.542 A)² × 30.0 Ω
= 0.293764 A² × 30.0 Ω
= 8.8129 W
≅ 8.813 W
iii. Current in 88.0 Ω
Current in 88.0 Ω is I = 0.542 A since the resistors are in series.
iv. Power in the 88.0 Ω
The power in the 88.0 Ω is P = I²R₂ where I = current = 0.542 A and R₂ = resistance = 88.0 Ω
So, P₂ = I²R₂
= (0.542 A)² × 88.0 Ω
= 0.293764 A² × 88.0 Ω
= 25.8512 W
≅ 25.85 W
(b) Repeat when the resistances are in parallel.
Since the resistors are connected in parallel, the same voltage is applied across them.
i. Current in 30.0 Ω
Using Ohm's law, V = I₁R₁ where V = voltage = 64.0 V, I₁ = current in 30.0 Ω resistor and R₁ = resistance = 30.0 Ω
So, I₁ = V/R₁ = 64.0 V/30.0 Ω = 2.13 A
ii Power in the 30.0 Ω
The power in the 30.0 Ω resistor is P₁ = V²/R₁ where V = voltage across resistor = 64.0 V and R₁ = resistance = 30.0 Ω
So, P₁ = V²/R₁
P₁ = (64.0 V)²/30.0 Ω
P₁ = 4096 V²/30.0 Ω
P₁ = 136.53 W
iii. Current in 88.0 Ω
Using Ohm's law, V = I₂R₂ where V = voltage = 64.0 V, I₂ = current in 88.0 Ω resistor and R₂ = resistance = 88.0 Ω
So, I₂ = V/R₂ = 64.0 V/88.0 Ω = 0.727 A
iv. Power in the 88.0 Ω
The power in the 30.0 Ω resistor is P₂ = V²/R₂ where V = voltage across resistor = 64.0 V and R₂ = resistance = 88.0 Ω
So, P₂ = V²/R₂
P₂ = (64.0 V)²/88.0 Ω
P₂ = 4096 V²/88.0 Ω
P₂ = 46.55 W
A 100-W light bulb is left on for 20.0 hours. Over this period of time, how much energy did the bulb use?
Answer:
Power = Energy/time
Energy = Power xtime.
Time= 20hrs
Power = 100Watt =0.1Kw
Energy = 0.1 x 20 = 2Kwhr.
This Answer is in Kilowatt-hour ...
If the one given to you is in Joules
You'd have to Change your time to seconds
Then Multiply it by the power of 100Watts.
A car hurtles off a cliff and crashes on the canyon floor below. Identify the system in which the net momentum is zero during the crash.
Solution :
It is given that a car ran off from a cliff and it crashes on canyon floor. Now the system of a car as well as the earth together have a [tex]\text{ net momentum of zero}[/tex] when the car crashes on the canyon floor, thus reducing the momentum of the car to zero. The earth also stops its upward motion and it also reduces the momentum to zero.
A boy is playing with a water hose, which has an exit area of
10 cm2 and has water flowing at a rate of 2 m/s. If he covers
the opening of the hose with his thumb so that it now has an
open area of 2 cm2, what will be the new exit velocity of the
water?
Answer:
The exit velocity of water is B. 15 m/s.
Explanation:
According to equation of continuity, for a steady flow of water, the volume of liquid entering a pipe in 1 second is equal to the volume that leaves per second.
If the initial exit area of the pipe is A₁ and the speed of exit is v₁ and the final exit area is A₂ and its corresponding exit velocity is v₂, then,
Rewrite the expression for v₂.
Substitute 10 cm² for A₁, 2 cm² for A₂ and 3 m/s for v₁.
The exit speed of water from the hose is 15 m/s.
A body of mass 4kg is moving with a velocity of 108km/h . find the kenetic energy of the body.
Answer:
KE = 2800 J
Explanation:
Usually a velocity is expressed as m/s. Then the energy units are joules.
[tex]\frac{108 km}{hr} * \frac{1000m}{1 km} * \frac{1 hour}{3600 seconds} =\frac{108*1000 m}{3600sec}[/tex]
v = 30 m / sec
KE = 1/2 * 4 * (30)^2
KE =2800 kg m^2/sec^2
KE = 2800 Joules
In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel to the x axis. At the instant particle A passes the y axis, particle B leaves the origin with zero initial speed and constant acceleration a with arrow of magnitude 0.35 m/s2. What angle between a with arrow and the positive direction of the y axis would result in a collision?
Answer:
59.26°
Explanation:
Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.
Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.
Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.
So, y = ut + 1/2a't²
y = 0 × t + 1/2(acosθ)t²
y = 0 + 1/2(acosθ)t²
y = 1/2(acosθ)t² (1)
Also, both particles must move the same horizontal distance to collide in time, t.
Let x be the horizontal distance,
x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision
Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.
So, x = ut + 1/2a"t²
x = 0 × t + 1/2(ainsθ)t²
x = 0 + 1/2(asinθ)t²
x = 1/2(asinθ)t² (3)
Equating (2) and (3), we have
vt = 1/2(asinθ)t² (4)
From (1) t = √[2y/(acosθ)]
Substituting t into (4), we have
v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²
v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)
v√[2y/(acosθ)] = ytanθ
√[2y/(acosθ)] = ytanθ/v
squaring both sides, we have
(√[2y/(acosθ)])² = (ytanθ/v)²
2y/acosθ = (ytanθ/v)²
2y/acosθ = y²tan²θ/v²
2/acosθ = ytan²θ/v²
1/cosθ = aytan²θ/2v²
Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1
secθ = ay(sec²θ - 1)/2v²
2v²secθ = aysec²θ - ay
aysec²θ - 2v²secθ - ay = 0
Let secθ = p
ayp² - 2v²p - ay = 0
Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have
ayp² - 2v²p - ay = 0
0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0
10.85p² - 15.68p - 10.85 = 0
dividing through by 10.85, we have
p² - 1.445p - 1 = 0
Using the quadratic formula to find p,
[tex]p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125[/tex]
Since p = secθ
secθ = 1.95625 or secθ = -0.51125
cosθ = 1/1.95625 or cosθ = 1/-0.51125
cosθ = 0.5112 or cosθ = -1.9956
Since -1 ≤ cosθ ≤ 1 we ignore the second value since it is less than -1.
So, cosθ = 0.5112
θ = cos⁻¹(0.5112)
θ = 59.26°
So, the angle between a with arrow and the positive direction of the y axis would result in a collision is 59.26°.
Based on the information in the table, what
is the acceleration of this object?
t(s) v(m/s)
0.0
9.0
1.0
4.0
2.0
-1.0
3.0
-6.0
A. -5.0 m/s2
B. -2.0 m/s2
C. 4.0 m/s2
D. 0.0 m/s2
Answer:
Option A. –5 m/s²
Explanation:
From the question given above, the following data were obtained:
Initial velocity (v₁) = 9 m/s
Initial time (t₁) = 0 s
Final velocity (v₂) = –6 m/s
Final time (t₂) = 3 s
Acceleration (a) =?
Next, we shall determine the change in the velocity and time. This can be obtained as follow:
For velocity:
Initial velocity (v₁) = 9 m/s
Final velocity (v₂) = –6 m/s
Change in velocity (Δv) =?
ΔV = v₂ – v₁
ΔV = –6 – 9
ΔV = –15 m/s
For time:
Initial time (t₁) = 0 s
Final time (t₂) = 3 s
Change in time (Δt) =?
Δt = t₂ – t₁
Δt = 3 – 0
Δt = 3 s
Finally, we shall determine the acceleration of the object. This can be obtained as follow:
Change in velocity (Δv) = –15 m/s
Change in time (Δt) = 3 s
Acceleration (a) =?
a = Δv / Δt
a = –15 / 3
a = –5 m/s²
Thus, the acceleration of the object is
–5 m/s².
A planet of mass m moves around the Sun of mass M in an elliptical orbit. The maximum and minimum distance of the planet from the Sun are r1 and r2, respectively. Find the relation between the time period of the planet in terms of r1 and r2.
Answer:
the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
Explanation:
Given the data i the question;
mass of sun = M
minimum and maximum distance = r1 and r2 respectively
Now, using Kepler's third law,
" the square of period T of any planet is proportional to the cube of average distance "
T² ∝ R³
average distance a = ( r1 + r2 ) / 2
we know that
T² = 4π²a³ / GM
T² = 4π² [( ( r1 + r2 ) / 2 )³ / GM ]
T² = 4π² [( ( r1 + r2 )³ / 8 ) / GM ]
T² = 4π² [( r1 + r2 )³ / 8GM ]
T = √[ 4π² [( r1 + r2 )³ / 8GM ] ]
T = 2π √[( r1 + r2 )³ / 8GM ]
Therefore, the relation between the time period of the planet is
T = 2π √[( r1 + r2 )³ / 8GM ]
