Answer:
[tex]{ \bf{v = u + at}} \\ 50 = 20 + (a \times 10) \\ 30 = 10a \\ { \tt{acceleration = 3 \: {ms}^{ - 2} }}[/tex]
When an automobile moves with constant velocity the power developed is used to overcome the frictional forces exerted by the air and the road. If the power developed in an engine is 50.0 hp, what total frictional force acts on the car at 55 mph (24.6 m/s)
P = F v
where P is power, F is the magnitude of force, and v is speed. So
50.0 hp = 37,280 W = F (24.6 m/s)
==> F = (37,280 W) / (24.6 m/s) ≈ 1520 N
(c) The ball leaves the tennis player's racket at a speed of 50 m/s and travels a
distance of 20 m before bouncing.
(i) Calculate how long it takes the ball to travel this distance.
(1 mark)
Answer:
t=0.417s
Explanation:
After the ball hits the racket it is in freefall(assume air resistance as negligible)
so a=-g
use
x-x0=v0t+1/2at^2
Plug in givens
20=50t-4.9t^2
Solve quadratic equation using quadratic formula
t= 0.417 seconds, (the other answer is extraneous because it is too big because in 1 second, the ball travels 50 meters)
Two guitar strings, of equal length and linear density, are tuned such that the second harmonic of the first string has the same frequency as the third harmonic of the second string. The tension of the first string is 510 N. Calculate the tension of the second string.
Answer:
The tension in the second string is 226.7 N.
Explanation:
Length is L, mass per unit length = m
T = 510 N
Let the tension in the second string is T'.
second harmonic of the first string = third harmonic of the second string
[tex]2 f = 3 f'\\\\2\sqrt{\frac{T}{m}} = 3 \sqrt {\frac{T'}{m}}\\\\4 T = 9 T'\\\\4\times 510 = 9 T'\\\\T' = 226.7 N[/tex]
If a bale of hay behind the target exerts a constant friction force, how much farther will your arrow burry itself into the hay than the arrow from the younger shooter
Answer:
The arrow will bury itself farther by 3S₁
Explanation:
lets assume; the Arrow shot by me has a speed twice the speed of the arrow fired by the younger shooter
Given that ; acceleration is constant , Frictional force is constant
A₂ = A₁
Vf²₂ - Vi²₂ / 2s₂ = Vf₁² - Vi₁² / 2s₁ ---- ( 1 )
final velocities = 0
Initial velocities : Vi₂ = 2(Vi₁ )
Back to equation 1
0 - (2Vi₁ )² / 2s₂ = 0 - Vi₁² / 2s₁
hence :
s₂ = 4s₁
hence the Arrow shot by me will burry itself farther by :
s₂ - s₁ = 3s₁
Note : S1 = distance travelled by the arrow shot by the younger shooter
Imagine you were given a converging lens and a meter stick and sent outside on a sunny day. In a few sentences, describe a method to measure, as accurately as possible, the focal length of the lens using only the lens, a meter stick, and your outside surroundings. Explain your reasoning
Answer:
the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance
Explanation:
The method for measuring the focal length of a lens is based on the use of the constructor's equation
[tex]\frac{1}{f } = \frac{1}{p} + \frac{1}{q}[/tex]
where q and q are the distance to the object and the image respectively, f is the focal length.
If we place the object very far away (infinity) the equation remains
[tex]\frac{1}{f} = \frac{1}{q}[/tex]
Therefore with this we can devise a means for measuring the Sun is the object that is at a very great distance and the focus point of the sun's shape is the distance to the magnesia, this image is equal to the seal distance
A study finds that the metabolic rate of mammals is proportional to m^3/4 , where m is the total body mass. By what factor does the metabolic rate of a 70.0-kg human exceed that of a 4.91-kg cat?
Answer:
The mass of human is 2898 times of the mass of cat.
Explanation:
A study finds that the metabolic rate of mammals is proportional to m^3/4 i.e.
[tex]M=\dfrac{km^3}{4}[/tex]
Where
k is constant
If m = 70 kg, the mass of human
[tex]M=\dfrac{70^3}{4}\\\\=85750[/tex]
If m = 4.91 kg, the mass of cat
[tex]M'=\dfrac{4.91^3}{4}\\\\=29.59[/tex]
So,
[tex]\dfrac{M}{M'}=\dfrac{85750}{29.59}\\\\=2897.93\approx 2898[/tex]
So, the mass of human is 2898 times of the mass of cat.
2. How do the phytochemicals present in various foods help us?
Phytochemicals are compounds that are produced by plants ("phyto" means "plant"). They are found in fruits, vegetables, grains, beans, and other plants. Some of these phytochemicals are believed to protect cells from damage that could lead to cancer.
The force an ideal spring exerts on an object is given by , where measures the displacement of the object from its equilibrium position. If , how much work is done by this force as the object moves from to
Answer:
The correct answer is "1.2 J".
Explanation:
Seems that the given question is incomplete. Find the attachment of the complete query.
According to the question,
x₁ = -0.20 mx₂ = 0 mk = 60 N/mNow,
⇒ [tex]W=\int_{x_1}^{x_2}F \ dx[/tex]
⇒ [tex]=\int_{x_1}^{x_2}-kx \ dx[/tex]
⇒ [tex]=-k \int_{-0.20}^{0}x \ dx[/tex]
By putting the values, we get
⇒ [tex]=-(60)[\frac{x^2}{2} ]^0_{-0.20}[/tex]
⇒ [tex]=-60[\frac{0}{2}-\frac{0.04}{2} ][/tex]
⇒ [tex]=1.2 \ J[/tex]
write the formulae of magnesium chloride and sodium sulfate
Answer:
Magnesium Chloride: MgCl2
Sodium Sulfate: Na2SO4
a vehicle start moving at 15m/s. How long will it take to stop at a distance of 15m?
Answer:
Explanation:
Speed= distance/time
Or time = distance/speed
According to your question
Speed=15m/s
and. Distance=1.2km. ,we must change kilometer in meter because given speed is in m/s
D= 1.2km = 1.2×1000m =1200meter
Time = distance/ speed
1200/15 =80second
Or. 1min and 20 sec will be your answer.
A charge Q exerts a 1.2 N force on another charge q. If the distance between the charges is doubled, what is the magnitude of the force exerted on Q by q
Answer:
0.3 N
Explanation:
Electromagnetic force is F= Kq1q2/r^2, where r is the distance between charges. If r is doubled then the force will be 1/4F which is 0.3 N.
The magnitude of the force exerted on Q by q when the distance between them is doubled is 0.3 N
Coulomb's law equationF = Kq₁q₂ / r²
Where
F is the force of attraction K is the electrical constant q₁ and q₂ are two point charges r is the distance apart Data obtained from the question Initial distance apart (r₁) = rInitial force (F₁) = 1.2 NFinal distance apart (r₂) = 2rFinal force (F₂) =? How to determine the final forceFrom Coulomb's law,
F = Kq₁q₂ / r²
Cross multiply
Fr² = Kq₁q₂
Kq₁q₂ = constant
F₁r₁² = F₂r₂²
With the above formula, we can obtain the final force as follow:
F₁r₁² = F₂r₂²
1.2 × r² = F₂ × (2r)²
1.2r² = F₂ × 4r²
Divide both side by 4r²
F₂ = 1.2r² / 4r²
F₂ = 0.3 N
Learn more about Coulomb's law:
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A particle of mass 1.2 mg is projected vertically upward from the ground with a velocity of 1.62 x 10 cm/h. Use the above information to answer the following four questions: 7. The kinetic energy of the particle at time t = 0 s is A. 1.215 x 10-3 J B. 2.430 J C. 1215 J D. 9.72 x 106 J E. OJ (2)
Answer:
K = 0 J
Explanation:
Given that,
The mass of the particle, m = 1.2 mg
The speed of the particle, [tex]v=1.62\times 10\ cm/h[/tex]
We need to find the kinetic energy of the particle at time t = 0 s.
At t = 0 s, the particle is at rest, v = 0
So,
[tex]K=\dfrac{1}{2}mv^2[/tex]
If v = 0,
[tex]K=0\ J[/tex]
So, the kinetic energy of the particle at time t = 0 s is 0 J.
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of with the horizontal
Complete Question
A 12.5-m fire truck ladder is leaning against a wall. Find the distance d the ladder goes up the wall (above the fire truck) if the ladder makes an angle of
40° 16' with the horizontal.
Answer:
[tex]d=8.01m[/tex]
Explanation:
From the question we are told that:
Length of ladder [tex]l=12.5m[/tex]
Angle [tex]\theta=40° 16'=20.26 \textdegree[/tex]
Generally the Trigonometric equation for distance d it goes up the wall is mathematically given by
[tex]d=l sin \theta[/tex]
[tex]d=12.5 sin 40.26[/tex]
[tex]d=8.01m[/tex]
A 5 kg object is moving in one dimension along the x-axis with a speed of 2 m/s. An external impulse acts on the force causing the speed of the object to increase to 5 m/s. The impulse lasted for 3 s. What is the average net force (in N) exerted on the object
Answer:
The correct answer is "15 Kg.m/s".
Explanation:
Given values are:
Mass,
m = 5 Kg
Initial velocity,
u = 2 m/s
Final velocity,
v = 5 m/s
Now,
The magnitude of change in linear momentum will be:
= [tex]m\times (v - u)[/tex]
By substituting the values, we get
= [tex]5\times (5 - 2)[/tex]
= [tex]5\times 3[/tex]
= [tex]15 \ Kg.m/s[/tex]
) Efficiency of a lever is always less than hundred percent.
Yes. Because it opposes the law of friction
I hope this helps.
Explanation:
Please mark me brainliest
A copper wire 1.0 meter long and with a mass of .0014 kilograms per meter vibrates in two segments when under a tension of 27 Newtons. What is the frequency of this mode of vibration
Answer:
the frequency of this mode of vibration is 138.87 Hz
Explanation:
Given;
length of the copper wire, L = 1 m
mass per unit length of the copper wire, μ = 0.0014 kg/m
tension on the wire, T = 27 N
number of segments, n = 2
The frequency of this mode of vibration is calculated as;
[tex]F_n = \frac{n}{2L} \sqrt{\frac{T}{\mu} } \\\\F_2 = \frac{2}{2\times 1} \sqrt{\frac{27}{0.0014} }\\\\F_2 = 138.87 \ Hz[/tex]
Therefore, the frequency of this mode of vibration is 138.87 Hz
find out the odd one and give reason (length, volume, time, mass
Answer:
Time
Explanation:
The answer to the question is actually time. Time is not needed when you calculate the mass or volume of an object, a square, sphere, rectangle, or any other 3D shape. You must also calculate the length to know what numbers you will be multiplying by. The answer to the question is time.
SCALCET8 3.9.018.MI. A spotlight on the ground shines on a wall 12 m away. If a man 2 m tall walks from the spotlight toward the building at a speed of 1.7 m/s, how fast is the length of his shadow on the building decreasing when he is 4 m from the building
Answer:
The length of his shadow is decreasing at a rate of 1.13 m/s
Explanation:
The ray of light hitting the ground forms a right angled triangle of height H, which is the height of the building and width, D which is the distance of the tip of the shadow from the building.
Also, the height of the man, h which is parallel to H forms a right-angled triangle of width, L which is the length of the shadow.
By similar triangles,
H/D = h/L
L = hD/H
Also, when the man is 4 m from the building, the length of his shadow is L = D - 4
So, D - 4 = hD/H
H(D - 4) = hD
H = hD/(D - 4)
Since h = 2 m and D = 12 m,
H = 2 m × 12 m/(12 m - 4 m)
H = 24 m²/8 m
H = 3 m
Since L = hD/H
and h and H are constant, differentiating L with respect to time, we have
dL/dt = d(hD/H)/dt
dL/dt = h(dD/dt)/H
Now dD/dt = velocity(speed) of man = -1.7 m/s ( negative since he is moving towards the building in the negative x - direction)
Since h = 2 m and H = 3 m,
dL/dt = h(dD/dt)/H
dL/dt = 2 m(-1.7 m/s)/3 m
dL/dt = -3.4/3 m/s
dL/dt = -1.13 m/s
So, the length of his shadow is decreasing at a rate of 1.13 m/s
A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the vehicles get stuck together. What is their velocity after the collision? A. 2.9 m/s east B. 1.6 m/s east m C. 2.6 m/s east D. 1.8 m/s east
Answer:
Explanation:
This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is
[tex][m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a[/tex] Filling in:
[tex][1200(4.5)+2100(0)]=[(1200+2100)v][/tex] which simplifies to
5400 + 0 = 3300v
so v = 1.6 m/s to the east, choice B
A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 59.4 m/s and returns the shot with the ball traveling horizontally at 37.2 m/s in the opposite direction. (Take the direction of the ball's final velocity (toward the net) to be the +x-direction).
(a) What is the impulse delivered to the ball by the racket?
(b) What work does the racket do on the ball?
5 9 . 4
- 3 7 . 2
2 2 . 2
Explanation:
Use the algorithm method.
5 9 . 4
- 3 7 . 2
2 2 . 2
2 Therefore, 59.4-37.2=22.259.4−37.2=22.2.
22.2
22.2
An electric eel can generate a 180-V, 0.1-A shock for stunning its prey. What is the eel's power output
Power output = volts x amps
Power output = 170 volts x 0.1 amps
Power output = 18 watts
Explain whether the unit of work is a fundamental or derived unit
Answer:
answer here
Explanation:
the unit of work is fundamental unit because it doesn't depend on other units.
__________________
Thx
Answer:
The SI unit of work is joule (J)
Explanation:
Joule is a derived unit. ∴ unit of work is a derived unit
chemical kinetics half lives
You are driving to the grocery store at 20 m/s. You are 150 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.
Required:
a. How far are you from the intersection when you begin to apply the brakes?
b. What acceleration will bring you to rest right at the intersection?
c. How long does it take you to stop?
Hi there!
a.
Use the formula d = st to solve:
d = 20 × 0.5 = 10m
150 - 10 = 140m away when brakes are applied
b.
Use the following kinematic equation to solve:
vf² = vi² + 2ad
Plug in known values:
0 = 20² + 2(150)(a)
Solve:
0 = 400 + 300a
-300a = 400
a = -4/3 (≈ -1.33) m/s² required
c.
Use the following kinematic equation to solve:
vf = vi + at
0 = 20 - 4/3t
Solve:
4/3t = 20
Multiply both sides by 3/4 for ease of solving:
t = 15 sec
Cold air rises because it is denser than water, is this true?
Answer:
true
Explanation:
im not sure please dont attack me
If a car generates 22 hp when traveling at a steady 100 km/h , what must be the average force exerted on the car due to friction and air resistance
Answer:
The average force exerted on the car is 590.12 N.
Explanation:
Given that,
The power generated, P = 22 hp = 16405.4 W
Speed of the car, v = 100 km/h = 27.8 m/s
We need to find the average force exerted on the car due to friction and air resistance.
We know that,
Power, P = F v
Where
F is force exerted on the car
[tex]F=\dfrac{P}{v}\\\\F=\dfrac{16405.4}{27.8}\\\\F=590.12\ N[/tex]
So, the average force exerted on the car is 590.12 N.
In a double-slit experiment, the slit spacing is 0.120 mm and the screen is 2.00 m from the slits. Find the wavelength (in nm) if the distance between the central bright region and the third bright fringe on a screen is 2.75 cm.
Answer:
[tex]\lambda=550nm[/tex]
Explanation:
From the question we are told that:
The slit spacing [tex]d_s=0.120mm=>0.120*10^{-3}[/tex]
Screen distance [tex]d_{sc}=2.0m[/tex]
Third Distance [tex]X=2.75cm=>2.75*10^{-2}[/tex]
Generally the equation for Wavelength is mathematically given by
[tex]\lambda=\frac{Xd_s}{n*d_{sc}}[/tex]
Where
n=number of screens
[tex]n=3[/tex]
Therefore
[tex]\lambda=\frac{2.75*10^{-2}*0.120*10^{-3}}{3*2}[/tex]
[tex]\lambda=0.055*10^{-5}[/tex]
[tex]\lambda=550nm[/tex]
Match each term with the best description.
a. Tightly woven fabric used to smother and extinguish a fire.
b. Consists of absorbent material that can be ringed around a chemical spill until the spill can be neutralized.
c. Device used to control small fires in an emergency situation
d. Provides chemical. physical. Health, and safety information regarding chemical reagents and supplies
1. Spill containment kit
2. Safety Data sheet
3. Fume hood
4. Fire extinguisher
5. Fire blanket
Answer:
A - 5
B - 1
C - 4
D -2
Explanation:
I don't have one i just know...
The fire blanket is a tightly woven fabric. The spill containment kit consists of absorbent material. Fire extinguishers control small fires and the safety data sheet provides chemical, health, and safety information.
(a) The fire blanket is a blanket, which may be quickly thrown over a fire to snuff out the flames, and comprises fire-resistant materials.
Hence, option (a) matches with option (5)
(b) In order to contain a chemical spill, absorbent items like pads, socks, or booms are frequently included in spill containment kits.
Hence, option (b) correctly matches with option (1).
(c) A fire extinguisher is a tool used to put out small fires during emergencies.
Hence, option (c) correctly matches with option (4).
(d) A Safety Data Sheet (SDS) gives in-depth details regarding a specific chemical or chemical mixture. It provides information about the physical characteristics of the chemical, any potential risks, safe handling and storage practices, emergency response strategies, and more.
Hence, option (d) correctly matches option (2).
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A typical incandescent light bulb consumes 75 W of power and has a mass of 20 g. You want to save electrical energy by dropping the bulb from a height great enough so that the kinetic energy of the bulb when it reaches the floor will be the same as the energy it took to keep the bulb on for 2.0 hours. From what height should you drop the bulb, assuming no air resistance and constant g?
Answer:
h = 2755102 m = 2755.102 km
Explanation:
According to the given condition:
Potential Energy = Energy Consumed by Bulb
[tex]mgh = Pt\\\\h = \frac{Pt}{mg}[/tex]
where,
h = height = ?
P = Power of bulb = 75 W
t = time = (2 h)(3600 s/1 h) = 7200 s
m = mass of bulb = 20 g = 0.02 kg
g = acceleration due to gravity = 9.8 m/s²
Therefore,
[tex]h = \frac{(75\ W)(7200\ s)}{(0.02\ kg)(9.8\ m/s^2)}[/tex]
h = 2755102 m = 2755.102 km
A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on collision with a 0.300kg glider that is moving to the left with a speed of 2.27m/s . Suppose the collision is elastic.
Part A
Find the magnitude of the final velocity of the 0.160kg glider. m/s
Part B
Find the direction of the final velocity of the 0.160kg glider.
i. to the right
ii. to the left
Part C
Find the magnitude of the final velocity of the 0.300kg glider. m/s
Part D
Find the direction of the final velocity of the 0.300kg glider.
Answer:
A) v_{f1} = -3.2 m / s, B) LEFT , C) v_{f2} = -0.12 m / s, D) LEFT
Explanation:
This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.
Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s
subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s
Initial instant. Before the crash
p₀ = m₁ v₀₁ + m₂ v₀₂
Final moment. After the crash
p_f = m₁ v_{f1} + m₂ v_{f2}
p₀ = p_f
m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}
as the shock is elastic, energy is conserved
K₀ = K_f
½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁ [tex]v_{f1}^2[/tex] + ½ m₂ [tex]v_{f2}^2[/tex]
m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)
let's make the relationship
(a + b) (a-b) = a² -b²
m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
let's write our two equations
m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂) (1)
m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)
we solve
v₀₁ + v_{f2} = v_{f2} + v₀₂
we substitute in equation 1 and obtain
M = m₁ + m₂
[tex]v_{f1} = \frac{m_1-m_2}{M} v_o_1 + 2 \frac{m_2}{M} v_f_2[/tex]
[tex]v_f_2 = \frac{2m_1}{M} v_o_1 + \frac{m_2-m_1}{M} v_o_2[/tex]vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2
we calculate the values
m₁ + m₂ = 0.160 +0.3000 = 0.46 kg
v_{f1} = [tex]\frac{ 0.160 -0.300} {0.460} \ 0.820 + \frac{2 \ 0300}{0.460} \ (-2.27)[/tex]
v_{f1} = -0,250 - 2,961
v_{f1} = - 3,211 m / s
v_{f2} = [tex]\frac{2 \ 0.160}{0.460} \ 0.820 + \frac{0.300 - 0.160}{0.460 } \ (-2.27)[/tex]
v_{f2} = 0.570 - 0.6909
v_{f2} = -0.12 m / s
now we can answer the different questions
A) v_{f1} = -3.2 m / s
B) the negative sign indicates that it moves to the left
C) v_{f2} = -0.12 m / s
D) the negative sign indicates that it moves to the LEFT
