Ask question

Ask question

All categories

3 years ago

11

Gina is driving her car down the street. She has a teddy bear sitting on the back seat. A dog runs in front of Gina's car, so sh

e quickly applies the brakes. The force of the brakes causes the car to stop, but the teddy bear continues to move forward until it hits the car's dashboard. The teddy bear did not stop at the same time as the car because A. Objects in motion tend to stay in motion unless acted upon by an outside force. B. More force is required to stop softer objects than to stop harder objects. C. Only objects touching the Earth's surface can be acted upon by an outside force. D. Objects in motion can only be stopped by the application of a balanced force.

1 answer:

Elza [17]3 years ago

6 0

Answer:

A

Explanation:

A. Objects in motion tend to stay in motion unless acted upon by an outside force.

You might be interested in

A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is

Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

$\frac{\Delta l}{l_{o}}=\frac{F}{AY}$ (1)

$T=2 \pi \sqrt{\frac{l_{o}}{g}}$ (2)

Where:

$\Delta l=0.05 mm=5(10)^{-5} m$ is the length the steel wire streches (taking into account 1mm=0.001 m)

$l_{o}$ is the length of the steel wire before being streched

$F=mg=(2 kg)(9.8 m/s^{2})=19.6 N$ is the force due gravity (the weight) acting on the pendulum with mass $m=2 kg$

$A$ is the transversal area of the wire

$Y=2(10)^{11} Pa$ is the Young modulus for steel

$T$ is the period of the pendulum

$g=9.8 m/s^{2}$ is the acceleration due gravity

Knowing this, let's begin by finding $A$ :

$A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4}$ (3)

Where $d=1.1 mm=0.0011 m$ is the diameter of the wire

$A=\pi \frac{(0.0011 m)^{2}}{4}$ (4)

$A=9.5(10)^{-7}m^{2}$ (5)

Knowing this area we can isolate $l_{o}$ from (1):

$l_{o}=\frac{\Delta l AY}{F}$ (6)

And substitute $l_{o}$ in (2):

$T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}}$ (7)

$T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}}$ (8)

Finally:

$T=1.39 s$

3 0

3 years ago

2. A rock is dropped off a bridge. How fast is the rock

zhenek [66]

Answer:

a) The velocity of rock at 1 second, v = 9.8 m/s

b) The velocity of rock at 3 second, v = 29.4 m/s

c) The velocity of rock at 5.5 second, v = 53.9 m/s

Explanation:

Given data,

The rock is dropped from a bridge.

The initial velocity of the rock, u = 0

a) The velocity of rock at 1 second,

Using the first equation of motion

v = u + gt

v = 0 + 9.8 x 1

v = 9.8 m/s

b) The velocity of rock at 3 second,

v = u + gt

v = 0 + 9.8 x 3

v = 29.4 m/s

c) The velocity of rock at 5.5 second,

v = u + gt

v = 0 + 9.8 x 5.5

v = 53.9 m/s

5 0

3 years ago

What's a good way to determine the net force of something

ollegr [7]

By adding up all the individual forces of the object

3 0

3 years ago

Which statement about ammeters and voltmeters is true?

Morgarella [4.7K]

What are the staments

7 0

3 years ago

A jet liner must reach a speed of 82 m/s for takeoff. If the

SIZIF [17.4K]

Answer:

The acceleration that the jet liner that must have is 2.241 meters per square second.

Explanation:

Let suppose that the jet liner accelerates uniformly. From statement we know the initial ( $v_{o}$ ) and final speeds ( $v_{f}$ ), measured in meters per second, of the aircraft and likewise the runway length ( $d$ ), measured in meters. The following kinematic equation is used to calculate the minimum acceleration needed ( $a$ ), measured in meters per square second:

$a = \frac{v_{f}^{2}-v_{o}^{2}}{2\cdot d}$

If we know that $v_{o} = 0\,\frac{m}{s}$ , $v_{f} = 82\,\frac{m}{s}$ and $d = 1500\,m$ , then the acceleration that the jet must have is:

$a = \frac{\left(82\,\frac{m}{s} \right)^{2}-\left(0\,\frac{m}{s} \right)^{2}}{2\cdot (1500\,m)}$

$a = 2.241\,\frac{m}{s^{2}}$

The acceleration that the jet liner that must have is 2.241 meters per square second.

3 0

2 years ago