Find the mass of a 165 N child

A car moving at 50 miles per hour speeds up steadily to 70 miles per hour over a period of 30 minutes. How far did it travel dur

eimsori [14]

Answer:

The car will travel 30 miles during the 30-minutes period of acceleration.

Explanation:

Given data :

Initial velocity = v₁ = 50 miles/hour

Final velocity = v₂ = 70 miles/hour

Time = t = 30 min = 0.5 hour

Using the definition of acceleration, we find the acceleration (a)

a = (v₂ - v₁) ÷ t

a = (70 - 50) ÷ 0.5

a = 20 ÷ 0.5

a = 40 miles/hour²

Using 3rd equation of motion, we find the distance travel (s)

2as = v₂² - v₁²

2(40)s = 70² - 50²

80 × s = 4900 - 2500

s = 2400 ÷ 80

s = 30 miles

5 0

3 years ago

Single Replacement: HCI + Zn

Kay [80]

During a single replacement, the Zn and H will switch places. That means the product formed will be ZnCl and H2.

4 0

3 years ago

A 23.3-kg mass is attached to one end of a horizontal spring, with the other end of the spring fixed to a wall. The mass is pull

sdas [7]

In spring mass system we know that angular frequency is given as

$\omega = 2\pi f$

f = 8.38 Hz

$\omega = 2\pi(8.38)$

$\omega = 52.65 rad/s$

now we know that speed of SHM at its extreme position is given by

$v = A\omega$

here we know that

A = 17.5 cm

$v = 0.175 (52.65)$

$v = 9.21 m/s$

so maximum speed is 9.21 m/s

7 0

3 years ago

A ball of moist clay falls 17.3 m to the ground. It is in contact with the ground for 24.0 ms before stopping. (a) What is the a

gizmo_the_mogwai [7]

Answer:

Acceleration, $767.08\ m/s^2$

Explanation:

Given that,

Height from a ball falls the ground, h = 17.3 m

It is in contact with the ground for 24.0 ms before stopping.

We need to find the average acceleration the ball during the time it is in contact with the ground.

Firstly, find the velocity when it reached the ground. So,

$v^2=u^2+2ah$

u = initial velocity=0 m/s

a = acceleration=g

$v=\sqrt{2gh} \\\\v=\sqrt{2\times 9.8\times 17.3} \\\\v=18.41\ m/s$

It is in negative direction, u = -18.41 m/s

Let a is average acceleration of the ball. Consider, v = and u = -18.41 m/s.

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-18.41}{24\times 10^{-3}}\\\\a=767.08\ m/s^2$

So, the average acceleration of the ball during the time it is in contact is $767.08\ m/s^2$ .

4 0

3 years ago

What is an example of potential energy to kinetic energy?

olga_2 [115]

Potential energy is the store she energy from an object this could include rubber bands. Kinetic energy is the energy that deals with motion a good example is a person running

6 0

2 years ago

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