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yawa3891 [41]
3 years ago
15

An object takes 5s to reach the ground from a hight of 5m on planet.what is the value of g on planet?

Physics
1 answer:
Greeley [361]3 years ago
8 0
T=5s
S=5m
U=0s/m
Acceleration=m/s^2
S=ut+(1/2)(a)(5^2)
A=(25/2)=5
A=5/(25/2)
A=10/25
A=2/5
A=0.4
A=0.4m/s^2.
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Please write in complete sentences.
BabaBlast [244]

Answer:

The density of a material affects the speed that a wave will be transmitted through it. In general, the denser the transparent material, the more slowly light travels through it.

8 0
3 years ago
Two steamrollers begin 105 mm apart and head toward each other, each at a constant speed of 1.20 m/s. At the same instant, a fly
levacccp [35]

Answer: 109.4 mm

Explanation: <u>Distance</u> is a scalar quantity and it is the measure of how much path there are between two locations. It can be calculated as the product of velocity and time:  d = vt

The separation between the two steamrollers is 105 mm or 0.105 m. They collide to each other at the middle of the separation:

location of collision = \frac{0.105}{2} = 0.0525 m

To reach that point, both steamrollers will have spent

v=\frac{\Delta x}{t}

t=\frac{\Delta x}{v}

t=\frac{0.0525}{1.2}

t = 0.04375 s

The fly is travelling with speed of 2.5 m/s. So, at t = 0.04375 s:

d = 2.5*0.04375

d = 0.109375 m

Until it is crushed, the fly will have traveled 109.4 mm.

4 0
3 years ago
A small particle has charge -2.00 uC and mass 1.50×10-4 kg. It moves from point A, where the electric potential is V(A) = 200 V,
Svet_ta [14]

Answer:

v_b=6.13 m/s..

Explanation:

Since no external force is acting on the system.

Therefore, Total energy remains constant before and after.

So, Total energy of system= energy due to potential applied+kinetic energy

T.E=qV_a+\dfrac{1}{2}mv_a^2=qV_b+\dfrac{1}{2}mv_b^2\\\dfrac{1}{2}mv_b^2=\dfrac{1}{2}mv_a^2+q(V_a-V_b)\\

(Here v=velocity ,V=potential ,q=charge and m=mass).

Putting values .

We get,  v_b=6.13 m/s..

At point B charged particle is moving faster as compared to point A.

Hence, it is the required solution.

5 0
4 years ago
Two point charges are on the y axis. A 3.90-µC charge is located at y = 1.25 cm, and a -2.4-µC charge is located at y = −1.80 cm
ladessa [460]

Answer:

a) 1.6*10^6 V

b) 13.35*10^6 V

Explanation:

The electric potential at origin is the sum of the contribution of the two charges. You use the following formula:

V=k\frac{q_1}{r_1}+k\frac{q_2}{r_2}    (1)

q1 = 3.90µC = 3.90*10^-6 C

q2 = -2.4µC = -2.4*10^-6 C

r1 = 1.25 cm = 0.0125 m

r2 = -1.80 cm = -0.018 m

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

You replace all the parameters in the equation (1):

V=k[\frac{q_1}{r_1}+\frac{q_2}{r_2}]\\\\V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0125m}+\frac{-2.4*10^{-6}C}{0.018m}]=1.6*10^6V

hence, the total electric potential is approximately 1.6*10^6 V

b) For the coordinate (1.50 cm , 0) = (0.015 m, 0) you have:

r1 = 0.0150m - 0.0125m = 0.0025m

r2= 0.015m + 0.018m = 0.033m

Then, you replace in the equation (1):

V=(8.98*10^9Nm^2/C^2)[\frac{3.90*10^{-6}C}{0.0025m}+\frac{-2.4*10^{-6}C}{0.033m}]=13.35*10^6V

hence, for y = 1.50cm you obtain V = 13.35*10^6 V

4 0
3 years ago
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BigorU [14]

Answer:

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Explanation:

4 0
3 years ago
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