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EleoNora [17]
2 years ago
9

Quddus and Saadman start running together from the same point. Quddus runs at constant velocity 0.5ms−1. Saadman’s initial veloc

ity is 0 but he runs at constant acceleration 0.1ms−2. How long will it take for them to meet each other?
a. 13.33
b. 10s
c. 50s
d. 26.67
Physics
1 answer:
svp [43]2 years ago
8 0

Answer:

the answer is C.

Explanation:

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PLEASE HELP!!! <br><br> i’ll mark brainliest if you’re correct
dedylja [7]

\qquad \qquad\huge \underline{\boxed{\sf Answer}}

The net force acting on the block is ~

\qquad \sf  \dashrightarrow \: 10 + 5

\qquad \sf  \dashrightarrow \: 15 \:N

So, the Answer in the boxes will be ~

\boxed{ \sf15} \:  \boxed{ \sf N}

7 0
3 years ago
Sophie says that geologic maps do not matter because she gets no benefits from them. Why is Sophie wrong? a. She can see the ran
astra-53 [7]
The Answer is Option C 
I think...
Sorry If i am wrong...
4 0
3 years ago
Read 2 more answers
222<br> A<br> 4.<br> Rn<br> Y + *He<br> 86<br> Z<br> 2
marissa [1.9K]

Answer:

looking at this triggered my fight or flight.

Explanation:

5 0
3 years ago
During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi
MArishka [77]

Answer:

0.3814128 m

Explanation:

t = Time taken = 36 ms

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration = 60g

g = Acceleration due to gravity = 9.81 m/s²

v=u+at\\\Rightarrow u=v-at\\\Rightarrow u=0-(-9.81\times 60)\times 36\times 10^{-3}\\\Rightarrow u=21.1896\ m/s

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-21.1896^2}{2\times -9.81\times 60}\\\Rightarrow s=0.3814128\ m

The distance the person traveled is 0.3814128 m

4 0
3 years ago
En la Tierra un volcán puede expulsar rocas verticalmente hasta una altura máxima H. A) ¿A qué altura (en términos de H) llegarí
Nonamiya [84]

A) 2.64 H

The maximum height that the expelled rock can reach can be found by using the equation:

v^2-u^2 = 2gd

where

v = 0 is the velocity at the maximum height

u is the initial velocity

g is the acceleration of gravity

d is the maximum height

Solving for d,

d=\frac{-u^2}{2g}

We see that the maximum heigth is inversely proportional to g. On the Earth,

d=H and g=g_e = -9.81 m/s^2

So we can write:

\frac{H}{H'}=\frac{g_m}{g_e}

where H' is the maximum height reached on Mars, and g_m = -3.71 m/s^2 is the acceleration of gravity on Mars. Solving for H',

H' = \frac{g_e}{g_m}H = \frac{-9.81}{3.71}H=2.64 H

B) 2.64T

The time after which the rock reaches the maximum height can be found by using

v=u+gt

where

v = 0 is the velocity at the maximum height

u is the initial velocity

Solving for t,

t=\frac{v-u}{g}

The total time of the motion is twice this value, so:

t=2\frac{v-u}{g}

So we see that it is inversely proportional to g.

On the Earth, t = T. So we can write:

\frac{T}{T'}=\frac{g_m}{g_E}

where T' is the total time of the motion on Mars. Solving for T',

T' = \frac{g_e}{g_m}T=\frac{-9.81}{-3.71}T=2.64T

4 0
3 years ago
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