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lions [1.4K]
2 years ago
11

Micah knows that a car had a change in velocity of 15 m/s. What does micah need to determine acceleration.

Physics
2 answers:
katrin2010 [14]2 years ago
4 0
The time component is needed. The acceleration is the change of velocity divided by the time in when this change of velocity happens.
Nadusha1986 [10]2 years ago
3 0

Answer:

He needs time which is B.

Explanation:

I took the test

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Steel Usually forms a
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Answer:

Permanent magnetism (of the steel)

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Part B For this activity, you'll analyze the motion of the marked spot on the skateboarder's leg. For each dropdown labeled “cm”
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Answer:

Explanation:

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7 0
2 years ago
Which one of the following lines best illustrates personification?
Mrrafil [7]
The answer would be A, because the wind cannot complain, therefore it has been given a human quality.
7 0
3 years ago
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A force, F1, of magnitude 2.0 N and directed due east is exerted on an object. A second force exerted on the object is F2 = 2.0
IgorLugansk [536]

Answer:

Magnitude of the force is

F_3 = 2.83

direction of the force is given as

\theta = 45 degree West of South

Explanation:

As we know that force is a vector quantity and in order to find the resultant of two or more forces we need to add them vectorialy

So here we have

\vec F_1 + \vec F_2 + \vec F_3 = 0

here we know that first force is of magnitude 2 N towards east

\vec F_1 = 2 \hat i N

second force is also of 2.0 N due North

\vec F_2 = 2 \hat j

now from above equation

2\hat i + 2\hat j + \vec F_3 = 0

\vec F_3 = -2\hat i - 2\hat j

so magnitude of the force is given as

F_3 = \sqrt{2^2 + 2^2}

F_3 = 2.83

direction of the force is given as

\theta = tan^{-1}\frac{F_y}{F_x}

\theta = tan^{-1}\frac{-2}{-2}

\theta = 45 degree West of South

3 0
2 years ago
Read 2 more answers
You're driving down the highway late one night at 20 m/s when a deer steps onto the road 44 m in front of you. Your reaction tim
never [62]

Answer:

14.0 m

25.1 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

Distance traveled in the reaction time

Distance = Speed × Time

\text{Distance}=20\times 0.5=10\ m

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-20^2}{2\times -10}\\\Rightarrow s=20\ m

Distance in which the car will stop is 10+20 = 30.0 m

So, the car will not hit the deer

Distance between the car and deer is 44-30 = 14.0 m

\text{Distance}=u\times 0.5=0.5u\ m

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u^2=v^2-2as\\\Rightarrow u^2=0^2-2\times -10\times (44-0.5u)\\\Rightarrow u^2=20(44-0.5u)\\\Rightarrow u^2=880-10u\\\Rightarrow u^2+10u-880=0

u=\frac{-10+\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}, u=\frac{-10-\sqrt{10^2-4\cdot \:1\left(-880\right)}}{2\cdot \:1}\\\Rightarrow u=25.08, -35.08\ m/s

Maximum speed of the car by which it will not hit the deer is 25.1 m/s

5 0
2 years ago
Read 2 more answers
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