All categories

3 years ago

A car is moving north at 5.2 m/s2. Which type of motion do the SI units in this value express?

Physics

1 answer:

irina [24]3 years ago

7 0

Answer:

the SI unit (meter per second square) indicates a linear type of motion.

Explanation:

Given;

acceleration of the car, a = 5.2 m/s² North

the SI unit of the car, = m/s²

The SI unit of the given value (acceleration), indicates a linear type of motion.

Linear acceleration is the change in linear velocity with time. Also, the northwards direction indicates linear displacement of the car.

Therefore, the SI unit (meter per second square) indicates a linear type of motion.

You might be interested in

What unit is used when measuring the push that cause a charge to move

Veronika [31]

I think that the answer is D. Pressure
Hope This Helps :D

5 0

3 years ago

A green croquet ball of mass 0.50 kg is rolling at +12 m/s. It collides with a blue croquet ball that also

Svetach [21]

Answer:

a) 9.6 m/s

b) 11.7 m/s

c) 12 m/s

Explanation:

This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{g} V_{o} + m_{b} U_{o}$ (2)

$p_{f}=m_{g} V_{f} + m_{b} U_{f}$ (3)

$m_{g}=0.5 kg$ is the mass of green ball

$m_{b}=0.5 kg$ is the mass of the blue ball

$V_{o}=12 m/s$ is the initial velocity of the green ball

$U_{o}=0 m/s$ is the initial velocity of the blue ball

$V_{f}$ is the final velocity of the green ball

$U_{f}$ is the final velocity of the blue ball

Substituting (2) and (3) in (1):

$m_{g} V_{o} + m_{b} U_{o}=m_{g} V_{f} + m_{b} U_{f}$ (4)

Isolating $U_{f}$ :

$U_{f}=\frac{m_{g} V_{o} - m_{g} V_{f}}{m_{b}}$ (5)

$U_{f}=\frac{m_{g} (V_{o} - V_{f})}{m_{b}}$ (6) This is the equation we will use for the next cases

Knowing this, let's begin with the answers:

a) In this case $V_{f}=2.4 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 2.4 m/s)}{0.5 kg}$ (7)

$U_{f}=9.6 m/s$ (8)

b) In this case $V_{f}=0.3 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 0.3 m/s)}{0.5 kg}$ (9)

$U_{f}=11.7 m/s$ (10)

c) In this case $V_{f}=0 m/s$ and we have to find $U_{f}$

$U_{f}=\frac{0.5 kg (12 m/s - 0 m/s)}{0.5 kg}$ (11)

$U_{f}=12 m/s$ (12)

4 0

3 years ago

A young woman walks up 55 steps to the top of a water slide. She slides

navik [9.2K]

Answer:

potential energy PE = M g h

KE at bottom = 1/2 M V^2

Regardless of the slope of the slide the change in energy is the same

1/2 V^2 = g h

V = (2 g h)^1/2 = (2 * 9.8 m/s^2 * 10 m)^1/2 = 14 m / s

Perhaps the question says that h = 55 * .1 = 5.5 m

Then V = (2 * 9.8 * 5.5) = 10.4 m/s

5 0

2 years ago

A 2kg ball is thrown with an acceleration of 15m/s2. A 2kg ball is thrown with an acceleration of 10m/s2. Which ball

DerKrebs [107]

A :-) for this question , we should apply
F = ma
( i ) Given - m = 2 kg
a = 15 m/s^2
Solution :
F = ma
F = 2 x 15
F = 30 N

( ii ) Given - m = 2 kg
a = 10 m/s^2
Solution :
F = ma
F = 2 x 10
F = 20 N

.:. The net force of object ( i ) has greater force compared to object ( ii ) by
( 30 - 20 ) 10 N

5 0

3 years ago

How much mass should be attached to a vertical ideal spring having a spring constant (force constant)of 39.5 N/m so that it will

nordsb [41]

Answer:

m = 1 kg

Explanation:

Given that,

The force constant of the spring, k = 39.5 N/m

The frequency of oscillation, f = 1 Hz

The frequency of oscillation is given by the formula as formula as follows :

$f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}} \\\\f^2=\dfrac{k}{4m\pi^2}\\\\m=\dfrac{k}{4\pi^2 f^2}\\\\m=\dfrac{39.5}{4\pi^2 \times (1)^2}\\\\m=1\ kg$

So, the mass that is attached to the spring is 1 kg.

6 0

3 years ago