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3 years ago

11

Calculate the acceleration of a bus whose speed changes from 7 m/s to 16 m/s over a period of 5 s.

1 answer:

Bezzdna [24]3 years ago

7 0

Let’s do this together!

Okay so the acceleration formula is vf-vi over time .

So the initial velocity (vi) 7m/s final velocity (vf) is 16m/s so we’re going to subtract 16-7 which is 9
M/s

So the time is 5s so 9m/s divided into 5s is 1.8m/s/2

So the answer is 1.8m\s2

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Which of Newton's laws of motion describes the motion of an object that has a net<br> force of ON?

algol [13]

Newton's first and second laws of motion both do, but I think the one you're looking for is: <em>The First Law of Motion</em>. That description is a little more direct.

It says that if an object is not acted on by a net external force, then it continues in "constant, uniform motion".

3 0

2 years ago

A student moves a box across the floor by exerting 56.7 N of force and doing 195 J of

Paha777 [63]

Answer:

A. 3.4 m

Explanation:

Given the following data;

Force = 56.7N

Workdone = 195J

To find the distance

Workdone is given by the formula;

$Workdone = force * distance$

Making "distance" the subject of formula, we have;

$Distance = \frac {workdone}{force}$

Substituting into the equation, we have;

$Distance = \frac {195}{56.7}$

Distance = 3.4 meters.

8 0

3 years ago

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally

aivan3 [116]

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m

speed at which the ball strike = ?

vertical velocity = 0 m/s

time at which the ball strike

$s = \dfrac{1}{2}gt^2$

$t = \sqrt{\dfrac{2s}{g}}$

$t = \sqrt{\dfrac{2\times 100}{9.8}}$

t = 4.53 s

vertical velocity at the time 4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = $\dfrac{65}{4.53}$ =14.35 m/s

speed of the ball = $\sqrt{44.39^2+14.35^2}$

= 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

5 0

3 years ago

. Friction is a rubbing force that ___________ a spinning yo-yo.

Advocard [28]

The yo-yo speeds up when you rub it

3 0

3 years ago

Read 2 more answers

You place an ice cube of mass 7.50×10−3kg and temperature 0.00∘C on top of a copper cube of mass 0.540 kg. All of the ice melts,

lbvjy [14]

Answer:

The value is $T_c = 12 .1 ^oC$

Explanation:

From the question we are told that

The mass of the ice cube is $m_i = 7.50 *10^{-3} \ kg$

The temperature of the ice cube is $T_i = 0^o C$

The mass of the copper cube is $m_c = 0.540 \ kg$

The final temperature of both substance is $T_f = 0^oC$

Generally form the law of thermal energy conservation,

The heat lost by the copper cube = heat gained by the ice cube

Generally the heat lost by the copper cube is mathematically represented as

$Q = m_c * c_c * [T_c - T_f ]$

The specific heat of copper is $c_c = 385J/kg \cdot ^oC$

Generally the heat gained by the ice cube is mathematically represented as

$Q_1 = m_i * L$

Here L is the latent heat of fusion of the ice with value $L = 3.34 * 10^{5} J/kg$

So

$Q_1 = 7.50 *10^{-3} * 3.34 * 10^{5}$

=> $Q_1 = 2505 \ J$

So

$2505 = 0.540 * 385 * [T_c - 0 ]$

=> $T_c = 12 .1 ^oC$

4 0

3 years ago