Ask question

Ask question

All categories

3 years ago

11

A car is moving at 20 m/s when the driver sees a stop sign and comes to rest at 65m. How long does the car take to come to rest?

1 answer:

tatiyna3 years ago

7 0

65/20 = 3.25 seconds.

You might be interested in

Adding heat to a liquid causes which of the following physical changes?

SashulF [63]

In almost every case in nature, adding heat to a liquid
causes the density of the liquid to decrease. That is,
when the liquid gets warmer, it expands and occupies
more space.

The one big exception to this rule is water !

Starting with a block of ice at zero°C (32°F), as the ice melts,
becomes water at zero°C, and all the way to 4°C (about 39°F),
its density increases all the way. That is, it shrinks and occupies
less volume as it goes from ice at zero°C to water at 4°C.

This sounds like an interesting but insignificant quirk ... until
you realize that if water didn't do this, then life on Earth would
be impossible !

8 0

3 years ago

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is:

$m_{S}\cdot v_{S} = (m_{S}+m_{C})\cdot v -m_{C}\cdot v_{C}$

$v_{S} = \frac{(m_{S}+m_{C})\cdot v - m_{C}\cdot v_{C}}{m_{S}}$

$v_{S} = \left(1+\frac{m_{C}}{m_{S}} \right)\cdot v - \frac{m_{C}}{m_{S}} \cdot v_{C}$

$v_{S} = v +\frac{m_{C}}{m_{S}}\cdot (v-v_{C})$

$v_{S} = -15\,\frac{m}{s}+\left(\frac{2000\,kg}{2200\,kg} \right) \cdot \left(-15\,\frac{m}{s}-45\,\frac{m}{s} \right)$

$v_{S} = -69.545\,\frac{m}{s}$

The semi truck travels at an initial speed of 69.545 meters per second downwards.

3 0

2 years ago

A mother and her 35.0 -kg child are riding an escalator to the third level of a shopping mall. If the child's gravitational pote

notka56 [123]

The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

The potential energy increases by 3773 J

PE₂-PE₁=mg(h₂-h₁)

3773 J = 35.0 × 9.81 × (h₂-h₁)

(h₂-h₁) = 10.98

Case 2 ;

ΔPE =?

ΔPE=mg(h₂-h₁)

ΔPE=56.0 × 9.81 ×10.98

ΔPE=6031.97 J.

Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

To learn more about the gravitational potential energy, refer;

brainly.com/question/3884855#SPJ1

#SPJ1

8 0

1 year ago

Light of wavelength 600 nm passes though two slits separated by 0.22mm and is observed on a screen 1.1m behind the slits. The lo

Ne4ueva [31]

Answer:

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

Explanation:

We have to take into account the expression for the position of the fringes

$dsen\theta=m\lambda\\y=\frac{m\lambda D}{d}$

where m is the number of the maximum, d is the separation of the slits, D is the distance to the screen.

(a) By replacing we obtain

$y(m=1)=\frac{(1)(600*10^{-9}m)(1.1m)}{0.22m}=3*10^{-6}m\\\\y(m=1)=\frac{(-1)(600*10^{-9}m)(1.1m)}{0.22m}=-3*10^{-6}m$

(b) more information is required to solve this point. Please complete the information.

HOPE THIS HELPS!

4 0

3 years ago

Read 2 more answers

The work done by the brakes during braking is equal to

kow [346]

Answer:

The amount of pressure delivered to them

Explanation:

6 0

2 years ago