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3 years ago

So I'm struggling with rearranging kinematic formulas. Does anyone have any steps or something to help.

1 answer:

7 0

Rearranging formulas is all about simple algebra rules. Just like when solving for x in an equation, you're just isolating whichever variable you want. I'll work this one out for you and hopefully it'll help, but if you need more explanation, then feel free to comment!

D = ViT + 0.5at² Subtract ViT from both sides

D - ViT = 0.5at² Divide both sides by 0.5t²

$\frac{D - ViT}{0.5t^{2} } = \frac{0.5at^{2} }{0.5t^{2} }$ I wrote this step out a little more to show how your fraction will cancel

$\frac{D - ViT}{0.5t^{2} }$ = a I like to flip these around so the single variable is on the right

a = $\frac{D - ViT}{0.5t^{2} }$

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Blank is how high or low you think a sound is

vfiekz [6]

<span>The word is "pitch", which is exactly that: How "high" or "low" a sound is.</span>

3 0

3 years ago

Read 2 more answers

Use the impulse-momentum theorem to find how long a stone falling straight down takes to increase its speed from 3.7 m/s to 9.90

11111nata11111 [884]

<h2>Time taken is 0.632 seconds</h2>

Explanation:

Impulse momentum theorem is change in momentum is impulse.

Change in momentum = Impulse

Final momentum - Initial momentum = Impulse

Mass x Final velocity - Mass x Initial Velocity = Force x Time

Mass x Final velocity - Mass x Initial Velocity =Mass x Acceleration x Time

Final velocity - Initial Velocity = Acceleration x Time

Final velocity = 9.9 m/s

Initial Velocity = 3.7 m/s

Acceleration = 9.81 m/s²

Substituting

9.9 - 3.7 = 9.81 x Time

Time = 0.632 seconds

Time taken is 0.632 seconds

8 0

2 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of <u>angular momentum conservation</u>, there should be no external torque before and after

As the <u>asteroid is travelling directly towards the center of the Earth</u>, after impact ,it <u>does not impose any torque on earth's rotation,</u> So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid

⇒ $\frac{2}{5} \times M\times R^{2} \times W_{1}=}(\frac{2}{5} \times M\times R^{2}$ $+$ $M_{1}\times R^{2})\times(\frac{4}{5} \times W_{1})$

$\frac{1}{2} \times M\times R^{2}$ = $(\frac{2}{5} \times M\times R^{2}$ + $M_{1}\times R^{2})$

$M_{1}\times R^{2}=$ $\frac{1}{10} \times M\times R^{2}$

⇒ $M_{1}=}$ $\frac{1}{10} \times M$

3 0

2 years ago

If you push a crate across a factory floor at constant speed in a constant direction, what is the magnitude of the force of fric

poizon [28]

Answer:

The magnitude of the force of friction equals the magnitude of my push

Explanation:

Since the crate moves at a constant speed, there is no net acceleration and thus, my push is balanced by the frictional force on the crate. So, the magnitude of the force of friction equals the magnitude of my push.

Let F = push and f = frictional force and f' = net force

F - f = f' since the crate moves at constant speed, acceleration is zero and thus f' = ma = m (0) = 0

So, F - f = 0

Thus, F = f

So, the magnitude of the force of friction equals the magnitude of my push.

3 0

2 years ago

A student pulls a 50-newton sled with a force having a magnitude of 15 newtons. What is the magnitude of the force that the sled

Simora [160]

Answer:

Force = 35 N

Explanation:

From Newton's third law of motion, the boy must apply a force greater than the weight of the sled to lift it.

weight of sled = mg

where m is its mass and g the force of gravity on it.

weight of sled = 50 N

Force applied by the boy on the sled = 15 N

Since the force applied on the sled by the boy is lesser than the weight of the sled, then;

Force that the sled exerts on the student = 50 - 15

= 35 N

The force exerted by the sled on the student is 35 N.

5 0

3 years ago