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MrRissso [65]
3 years ago
8

Which of the following expressions uses the correct conversion factor to convert 35.7 km into the equivalent distance in miles?A

) 35.7 km × 1 km/1.609 milesB) 35.7 km × 1 mile/1.609 milesC) 35.7 km × 1 km/1.609 kmD) 35.7km × 1 mile/1.609 km
Physics
1 answer:
Lostsunrise [7]3 years ago
8 0

Answer: D) 35.7km × 1 mile/1.609 km

Explanation:

Given that;

We need to convert 35.7km to miles

And we know that 1.609km makes 1 mile = 1.609km/1mile

To convert it to mile we need to obtain the number of miles per kilometre

= 1/(1.609km/mile)

= 1mile/1.609km

Then we can now multiply the conversion rate by the number.

= 35.7km × 1mile/1.609km

= 22.19 miles

Note that the sign must cancel out to give miles.

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A ball of 0.1kg is thrown straight up into the air with
Contact [7]

a) The momentum of the ball at the maximum height is zero

b) The momentum of the ball at halfway is 1.06 kg m/s

Explanation:

a)

There is only one force acting on the ball during its motion: the force of gravity, acting downward. As a result, the ball hasa constant acceleration downward (g=9.8 m/s^2, acceleration of gravity), and therefore it is in free fall motion.

This means that the velocity of the ball is given by the equation:

v=u-gt

where

v is the velocity at time t

u = 15 m/s is the initial velocity

g=9.8 m/s^2

As we see from the equation, the term -gt is negative: this means that as the ball goes higher and higher, its velocity decreases. The point of maximum height is reached when the velocity has become zero (after that point, the velocity changes direction and points downward), so when

v=0

The momentum of the ball at any point is given by

p=mv

where m = 0.1 kg is its mass. Therefore, at the maximum height, since v=0, the momentum is also zero:

p=0

b)

We can use the following suvat equation to find the maximum height reached by the ball:

v^2-u^2=2as

where

v = 0 is the velocity at the point of maximum height

u = 15 m/s is the initial velocity

a=-g=-9.8 m/s^2 is the acceleration (downward)

s is the vertical displacement (the maximum height)

Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{0-(15)^2}{2(-9.8)}=11.48 m

This means that the ball is halfway when its height is

h'=\frac{s}{2}=\frac{11.48}{2}=5.74 m

We can find the velocity of the ball v' when it is at h' by using again the same equation:

v'^2-u^2=2ah'\\v=\sqrt{u^2+2ah'}=\sqrt{(15)^2+2(-9.8)(5.74)}=10.6 m/s

And therefore, the momentum of the ball here is

p'=mv'=(0.1)(10.6)=1.06 kg m/s

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

brainly.com/question/9484203

#LearnwithBrainly

7 0
3 years ago
If a fox starts from rest and accelerates at a rate of 1.1 m/s^2, how long does it take the fox to run 5 m A. 3.0s B. 4.5s C. 9.
KATRIN_1 [288]
Given:
u=0 m/s
a=1.1 m/s^2
S=5 m
t=time it takes to run 5 m

Use the kinematics equation
S=ut+(1/2)at^2
=>
5=0*t+(1/2)1.1(t^2)
solve for t
t=sqrt(5*2/1.1)=3.015 seconds.

8 0
4 years ago
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(D) Negative acceleration (less and less speed)
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Answer:

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7 0
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