(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is
W = - (1/2 kx ² - 1/2 k (0.0400 m)²)
(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)
By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so
W = ∆K = 0 - 1/2 m (0.835 m/s)²
Solve for x :
- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²
==> x ≈ 0.0493 m ≈ 4.93 cm
(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is
W = - 1/2 k (0.0400 m)²
If v is the glider's maximum speed, then by the work-energy theorem,
W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²
Solve for v :
- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²
==> v ≈ 1.43 m/s
(c) The angular frequency of the glider's oscillation is
√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz
The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s
From the given information;
the mass of the glider = 190 gForce constant k = 160 N/mthe horizontal speed of the glider [tex]v_x[/tex] = 0.835 m/sthe distance away from the equilibrium = 4.0 cm = 0.04 mUsing energy conservation E, the amplitude of the motion can be calculated by using the formula:
[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]
[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]
[tex]\mathbf{E =0.194 \ J}[/tex]
Similarly, we know that:
[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]
Making amplitude A the subject, we have:
[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]
[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]
[tex]\mathbf{A =0.049 \ cm}[/tex]
Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:
[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]
[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]
[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]
The angular frequency of the oscillation can be computed by using the expression:
[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]
[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]
ω = 29.02 rad/s
Learn more about energy conservation here:
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Một mặt phẳng vô hạn tích điện đều, mật độ σ = 4.10-9 C/cm2 , đặt thẳng đứng trong không khí. Một quả cầu nhỏ có khối lượng 8 g, mang điện tích q = 10-8 C treo gần vào mặt phẳng, sao cho dây treo lúc đầu song song với mặt phẳng. Lấy g = 9,8m/s2 . Khi cân bằng, dây treo quả cầu hợp với mặt phẳng 1 góc bằng bao nhiêu?
Answer:
The angle is 16 degree.
Explanation:
A uniformly charged infinite plane, density σ = 4.10-9 C/cm2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2 . In equilibrium, by what angle does the string hanging from the ball make an angle with the plane?
Surface charge density, σ = 4 x 10^-5 C/m^2
charge, q = 10^-8 C
mass, m = 0.008 kg
The electric field due to the plate is
[tex]E= \frac{\sigma }{2\varepsilon 0}[/tex]
Let the angle make with the vertical is A and T is the tension in the string.
[tex]T sin A = q E....(1)\\\\T cos A = m g .... (2)\\\\Divie (1) by (2)\\\\tan A =\frac{q E}{m g}\\\\tan A = \frac{10^{-8}\times 4\times 10^{-5}}{2\times 8.85\times 10^{-12}\times 0.008\times9.8}\\\\tan A = 0.288\\\\A = 16 degree\\[/tex]
Which of the following represents the velocity time relationship for a falling apple?
Answer "a" would be correct.
Answer:
d
Explanation:
There's an acceleration from gravity, thus the velocity is becoming faster and faster as it reaches the ground. Thus its D
Brainliest please~
Cell phone conversations are transmitted by high-frequency radio waves. Suppose the signal has wavelength 35 cm while traveling through air. What are the
(a) frequency and
(b) wavelength as the signal travels through 3-mm-thick window glass into your room?
Answer:
(a) 8.57 x 10^8 Hz
(b) 23.3 cm
Explanation:
Wavelength = 35 cm = 0.35 m
speed =3 x10^8 m/s
Let the frequency is f.
(a) The relation is
speed = frequency x wavelength
3 x 10^8 = 0.35 x f
f = 8.57 x 10^8 Hz
(b) refractive index of glass is 1.5
The relation for the refractive index and the wavelength is
wavelength in glass= wavelength in air/ refractive index.
Wavelength in glass= 35/1.5 = 23.3 cm
The following two waves are sent in opposite directions on a horizontal string so as to create a standing wave in a vertical plane: y1(x, t) = (8.20 mm) sin(4.00πx - 430πt) y2(x, t) = (8.20 mm) sin(4.00πx + 430πt), with x in meters and t in seconds. An antinode is located at point A. In the time interval that point takes to move from maximum upward displacement to maximum downward displacement, how far does each wave move along the string?
Answer:
Explanation:
From the information given:
The angular frequency ω = 430 π rad/s
The wavenumber k = 4.00π which can be expressed by the equation:
k = ω/v
∴
4.00 = 430 /v
v = 430/4.00
v = 107.5 m/s
Similarly: k = ω/v = 2πf/fλ
We can say that:
k = 2π/λ
4.00 π = 2π/λ
wavelength λ = 2π/4.00 π
wavelength λ = 0.5 m
frequency of the wave can now be calculated by using the formula:
f = v/λ
f = 107.5/0.5
f = 215 Hz
Also, the Period(T) = 1/215 secs
The time at which particle proceeds from point A to its maximum upward displacement and to its maximum downward displacement can be computed as t = T/2;
Thus, the distance(x) covered by each wave during this time interval(T/2) will be:
x = v * t
x = v * T/2
x = λ/2
x = 0.5/2
x = 0.25 m
what is time taken by radio wave to go and return back from communication satellite to earth??
Answer:
Radio waves are used to carry satellite signals. These waves travel at 300,000 km/s (the speed of light). This means that a signal sent to a satellite 38,000 km away takes 0.13 s to reach the satellite and another 0.13 s for the return signal to be received back on Earth.
Explanation:
hope it help
A mass-spring system oscillates with an amplitude of 4.20 cm. If the spring constant is 262 N/m and the mass is 560 g, determine the mechanical energy of the system.
Answer:
[tex]M.E=41J[/tex]
Explanation:
From the question we are told that:
Amplitude [tex]a=4.20cm[/tex]
Spring Constant [tex]K=262N/m[/tex]
Mass [tex]m=560g[/tex]
Generally the equation for mechanical energy is mathematically given by
[tex]M.E=\frac{1}{2}km^2[/tex]
[tex]M.E=0.5*262*0.56^2[/tex]
[tex]M.E=41J[/tex]
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.
190 students sit in an auditorium listening to a physics lecture. Because they are thinking hard, each is using 125 W of metabolic power, slightly more than they would use at rest. An air conditioner with a COP of 5.0 is being used to keep the room at a constant temperature. What minimum electric power must be used to operate the air conditioner?
Answer:
W = 4.75 KW
Explanation:
First, we will calculate the heat to be removed:
Q = (No. of students)(Metabolic Power of Each Student)
Q = (190)(125 W)
Q = 23750 W = 23.75 KW
Now the formula of COP is:
[tex]COP = \frac{Q}{W}\\\\W = \frac{Q}{COP}\\\\W = \frac{23.75\ KW}{5}\\\\[/tex]
W = 4.75 KW
g A mass of 2.0 kg traveling at 3.0 m/s along a smooth, horizontal plane hits a relaxed spring. The mass is slowed to zero velocity when the spring has been compressed by 0.15 m. What is the spring constant of the spring
By the work-energy theorem, the total work done on the mass by the spring is equal to the change in the mass's kinetic energy:
W = ∆K
and the work done by a spring with constant k as it gets compressed a distance x is -1/2 kx ²; the work it does is negative because the restoring force of the spring points opposite the direction in which it's getting compressed.
So we have
-1/2 k (0.15 m)² = 0 - 1/2 (2.0 kg) (3.0 m/s)²
Solve for k to get k = 800 N/m.
A point charge of -3.0 x 10-5C is placed at the origin of coordinates. Find the electric field at the point 3. r= 50 m on the x-axis
Answer: -5×10-3
Explanation:
E=kq/r
1.a machine gun fires a ball with an initial velocity of 600m/s with an elevation of 30° with respect to the ground neglecting air resistance calculate:
a.the maximum height that can be reached?
b.the time of flight of the bullet?
c.the maximum horizontal displacement of the ired bullet?
Answer:
See explanation
Explanation:
a) maximum height of a projectile = u sin^2θ/2g
H= 600 × (sin 30)^2/2 × 10
H= 7.5 m
b) Time of flight
t= 2u sinθ/g
t= 2 × 600 sin 30/10
t= 60 seconds
Range
R= u^2sin2θ/g
R= (600)^2 × sin2(30)/10
R= 31.2 m
g A spherical container of inner diameter 0.9 meters contains nuclear waste that generates heat at the rate of 872 W/m3. Estimate the total rate of heat transfer from the container to its surroudings ignoring radiation.
Answer: The total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.
Explanation:
Given: Inner diameter = 0.9 m
q = 872 [tex]W/m^{3}[/tex]
Now, radii is calculated as follows.
[tex]r = \frac{diameter}{2}\\= \frac{0.9}{2}\\= 0.45 m[/tex]
Hence, the rate of heat transfer is as follows.
[tex]Q = q \times V[/tex]
where,
V = volume of sphere = [tex]\frac{4}{3} \pi r^{3}[/tex]
Substitute the values into above formula as follows.
[tex]Q = q \times \frac{4}{3} \pi r^{3}\\= 872 W/m^{3} \times \frac{4}{3} \times 3.14 \times (0.45 m)^{3}\\= 332.67 W[/tex]
Thus, we can conclude that the total rate of heat transfer from the container to its surroundings ignoring radiation is 332.67 W.
The weight of a hydraulic barber's chair with a client is 2100 N. When the barber steps on the input piston with a force of 44 N, the output plunger of a hydraulic system begins to lift the chair. Determine the ratio of the radius of the output plunger to the radius of the input piston.
Answer:
[tex]\frac{r_1}{r_2}=6.9[/tex]
Explanation:
According to Pascal's Law, the pressure transmitted from input pedal to the output plunger must be same:
[tex]P_1 = P_2\\\\\frac{F_1}{A_1}=\frac{F_2}{A_2}\\\\\frac{F_1}{F_2}=\frac{A_1}{A_2}\\\\\frac{F_1}{F_2}=\frac{\pi r_1^2}{\pi r_2^2}\\\\\frac{F_1}{F_2}=\frac{r_1^2}{r_2^2}[/tex]
where,
F₁ = Load lifted by output plunger = 2100 N
F₂ = Force applied on input piston = 44 N
r₁ = radius of output plunger
r₂ = radius of input piston
Therefore,
[tex]\frac{r_1^2}{r_2^2}=\frac{2100\ N}{44\ N}\\\\\frac{r_1}{r_2}=\sqrt{\frac{2100\ N}{44\ N}} \\\\\frac{r_1}{r_2}=6.9[/tex]
The thrust F of a screw propeller is known to depend upon the diameter d, Speed of advance v, fluid density e, revolution per second N, and the coefficient of viscosity M, of the fluid. Find the expression for F, in terms of the quantities
Answer:
[tex]{ \bf{F = { \tt{ \frac{4}{3} \pi {r}^{3}v gM}}}}[/tex]
A point charge of -3.0 x 10-C is placed at the origin of coordinates. Find the clectric field at the point 13. X= 5.0 m on the x-axis.
Answer:
-1.0778×10⁻¹⁰ N/C
Explanation:
Applying,
E = kq/r²................ equation 1
Where E = elctric field, q = charge, r = distance, k = coulomb's law
From the question,
Given: q = -3.0×10 C, r = 5.0 m
Constant: k = 8.98×10⁹ Nm²/C²
Substitute these values in equation 1
E = (-3.0×10)(8.98×10⁹)/5²
E = -1.0778×10⁻¹⁰ N/C
Hence the electric field on the x-axis is -1.0778×10⁻¹⁰ N/C
A 165-km-long high-voltage transmission line 2.00 cm in diameter carries a steady current of 1,015 A. If the conductor is copper with a free charge density of 8.50 1028 electrons per cubic meter, how many years does it take one electron to travel the full length of the cable?
Answer:
22.1 years
Explanation:
Since the current in the wire is I = nevA where n = electron density = 8.50 × 10²⁸ electrons/cm³ × 10⁶ cm³/m³= 8.50 × 10³⁴ electrons/m³, e = electron charge = 1.602 × 10⁻¹⁹ C, v = drift velocity of electrons and A = cross-sectional area of wire = πd²/4 where d = diameter of wire = 2.00 cm = 2 × 10⁻² m
Making v subject of the formula, we have
v = I/neA
So, v = I/neπd²/4
v = 4I/neπd²
Since I = 1,015 A, substituting the values of the other variables into the equation, we have
v = 4I/neπd²
v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π ×(2 × 10⁻² m)²]
v = 4(1,015 A)/[8.50 × 10³⁴ electrons/m³ × 1.602 × 10⁻¹⁹ C × π × 4 × 10⁻⁴ m²]
v = (1,015 A)/[42.779 × 10¹¹ electronsC/m]
v = 23.73 × 10⁻¹¹ m/s
v = 2.373 × 10⁻¹⁰ m/s
Since distance d = speed, v × time, t
d = vt
So, the time it takes one electron to travel the full length of the cable is t = d/v
Since d = distance moved by free charge = length of transmission line = 165 km = 165 × 10³ m and v = drift velocity of charge = 2.373 × 10⁻¹⁰ m/s
t = 165 × 10³ m/2.373 × 10⁻¹⁰ m/s
t = 69.54 × 10⁷ s
t = 6.954 × 10⁸ s
Since we have 365 × 24 hr/day × 60 min/hr × 60 s/min = 31536000 s in a year = 3.1536 × 10⁷ s
So, 6.954 × 10⁸ s = 6.954 × 10⁸ s × 1yr/3.1536 × 10⁷ s = 2.21 × 10 yrs = 22.1 years
It will take one electron 22.1 years to travel the full length of the cable
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 ]
Need ur help,,, :-[ :-{
...... ............ .. ..
Answer:
Graph B express the magnetic relationship of magnetic flux and electronic flow
Which physical phenomenon is illustrated by the fact that the prism has different refractive indices for different colors
Answer:
The incoming white light is composed of light of different colors,
Since these different colors have different refractive indices they are refracted at different angles from one another.
The output light is then separated by color creating a color spectrum.
Since n is greater for shorter wavelengths (violet colors) these wavelengths are refracted thru the larger angles.
What is the total surface charge qint on the interior surface of the conductor (i.e., on the wall of the cavity)
Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
A
cook
holds a 3.2 kg carton of milk at arm's length.
75.9
w
25,5 cm
What force FB must be exerted by the bi-
ceps muscle? The acceleration of gravity is
9.8 m/s2. (Ignore the weight of the forearm.)
Answer in units of N.
Answer:
Explanation:
From the given information:
From the rotational axis, the distance of the force of gravity is:
d_g = 25+5.0 cm
d_g = 30.0 cm
d_g = 30.0 × 10⁻² m
However, the relative distance of FB cos 75.9° from the axis is computed as:
d_B = 5.0 cm
d_B = 5.0 × 10⁻² m
The net torque rotational equilibrium = zero (0)
i.e.
[tex]\tau_g -\tau_B = 0 \\ \\ F_gd_g -F_gcos 75.9^0 d_B = 0 \\ \\ F_B = \dfrac{F_g d_g}{F_g cos 65.6} \\ \\ F_B = \dfrac{(3.2)(9.8)(30*10^{-2})}{(5.0*10^{-2} * cos 75.9)} \\ \\ \mathbf{F_B = 772.4 N}[/tex]
= 772.4 N
Thus, the force exerted = 1772.4 N
Three 15-Ω and two 25-Ω light bulbs and a 24 V battery are connected in a series circuit. What is the current that passes through each bulb?
1) 0.18 A
2) 0.25 A
3) 0.51 A
4) 0.74 A
5) The current will be 1.6 A in the 15-Ω bulbs and 0.96 A in the 25-Ω bulbs.
Answer:
I = 0.25 A
Explanation:
Given that,
Three 15 ohms and two 25 ohms light bulbs and a 24 V battery are connected in a series circuit.
In series combination, the equivalent resistance is given by :
[tex]R=R_1+R_2+R_3+....[/tex]
So,
[tex]R=15+15+15+25+25\\\\=95\ \Omega[/tex]
The current each resistor remains the same in series combination. It can be calculated using Ohm's law i.e.
V = IR
[tex]I=\dfrac{V}{R}\\\\I=\dfrac{24}{95}\\\\I=0.25\ A[/tex]
So, the current of 0.25 A passes through each bulb.
Differentiate between Scalar quantity and Vector quantity and give two examples each
Electrical resistance is a measure of resistance to the flow of _?____
Resistance is a measure of the opposition to current flow in an electrical circuit. Resistance is measured in ohms, symbolized by the Greek letter omega (Ω). Ohms are named after Georg Simon Ohm (1784-1854), a German physicist who studied the relationship between voltage, current and resistance.
Hope this helps!!!!
Answer:
electric current
Explanation:
The answer is electric current
* A ball is projected horizontally from the top of
a building 19.6m high.
a, How long when the ball take to hit the ground?
b, If the line joining the point of projection to
the point where it hits the ground is 45
with the horizontal. What must be the
initial velocity of the ball?
c,with what vertical verocity does the ball strike
the grounds? (9= 9.8 M152)
Explanation:
Given
Ball is projected horizontally from a building of height [tex]h=19.6\ m[/tex]
time taken to reach ground is given by
[tex]\text{Cosidering vertical motion}\\\Rightarrow h=ut+0.5at^2\\\Rightarrow 19.6=0+0.5\times 9.8t^2\\\Rightarrow t^2=4\\\Rightarrow t=2\ s[/tex]
(b) Line joining the point of projection and the point where it hits the ground makes an angle of [tex]45^{\circ}[/tex]
From the figure, it can be written
[tex]\Rightarrow \tan 45^{\circ}=\dfrac{h}{x}\\\\\Rightarrow x=h\cdot 1\\\Rightarrow x=19.6[/tex]
Considering horizontal motion
[tex]\Rightarrow x=u_xt\\\Rightarrow 19.6=u_x\times 4\\\Rightarrow u_x=4.9\ m/s[/tex]
(c) The vertical velocity with which it strikes the ground is given by
[tex]\Rightarrow v^2-u_y^2=2as\\\Rightarrow v^2-0=2\times 9.8\times 19.6\\\Rightarrow v=\sqrt{384.16}\\\Rightarrow v=19.6\ m/s[/tex]
Thus, the ball strikes with a vertical velocity of [tex]19.6\ m/s[/tex]
Explanation:
Given
Ball is projected horizontally from a building of height
time taken to reach ground is given by
(b) Line joining the point of projection and the point where it hits the ground makes an angle of
From the figure, it can be written
Considering horizontal motion
(c) The vertical velocity with which it strikes the ground is given by
Thus, the ball strikes with a vertical velocity of
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]
Why don’t you see tides ( like those of the ocean ) in your swimming pool ?
Which of the following is a noncontact force?
O A. Friction between your hands
O B. A man pushing on a wall
O C. Air resistance on a car
D. Gravity between you and the Sun
Answer:
Gravity between you and the sun
state the story of archimedes
Answer:
Archimedes was born about 287 BCE in Syracuse on the island of Sicily. He died in that same city when the Romans captured it following a siege that ended in either 212 or 211 BCE. One story told about Archimedes' death is that he was killed by a Roman soldier after he refused to leave his mathematical work.
A body starts from rest and accelerates uniformly at 5m/s. Calculate the time taken by the body to cover a distance of 1km
Answer:
20 seconds
Explanation:
We are given 2 givens in the first statement
v0=0 and a=5
And we are trying to find time needed to cover 1km or 1000m.
So we use
x-x0=v0t+1/2at²
Plug in givens
1000=0+2.5t²
solve for t
t²=400
t=20s
