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Lelu [443]
3 years ago
6

A taxi hurries with a constant speed of 60 miles per hour. How long will it take to travel a distance of 130 miles

Physics
1 answer:
Alecsey [184]3 years ago
3 0

Answer:

2.17 hrs

Explanation:

time = distance / velocity

We know that distance = 130 miles and velocity = 60 miles/h.

t = d / v

t = 130 miles / 60 miles/h

t = 2.17 hrs

It takes about 2.17 hours to travel a distance of 130 miles. Hope this helps, thank you !!

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A mechanic has a wrench where the hand grip is 0.40 m from the axis of the bolt. To apply a torque of 170 N m with this wrench w
Grace [21]

Answer:

4.3 * 10 N

Explanation:

To calculate torque, we multiply the distance from the pivot by the perpendicular (the part of the force that acts at right angles to the displacement vector) component of the force to the displacement vector from the pivot.

torque  = distance from pivot *  perpendicular force

170 Nm= 0.4 m * F

F = 425 N = 4.3 * 10 N rounded off to two significant figures

4 0
3 years ago
Suppose a baseball pitcher throws the ball to his catcher.
amm1812

a) Same

b) Same

c) Same

d) Throw the ball takes longer

e) F is larger when the ball is catched

Explanation:

a)

The change in speed of an object is given by:

\Delta v = |v-u|

where

u is the initial velocity of the object

v is the final velocity of the object

The change in speed is basically the magnitude of the change in velocity (because velocity is a vector, while speed is a scalar, so it has no direction).

In this problem:

- In situation 1 (pitcher throwing the ball), the initial velocity is

u = 0 (because the ball starts from rest)

while the final velocity is v, so the change in speed is

\Delta v=|v-0|=|v|

- In situation 2 (catcher receiving the ball), the initial velocity is now

u = v

while the final velocity is now zero (ball coming to rest), so the change in speed is

\Delta v =|0-v|=|-v|

Which means that the two situations have same change in speed.

b)

The change in momentum of an object is given by

\Delta p = m \Delta v

where

m is the mass of the object

\Delta v is the change in velocity

If we want to compare only the magnitude of the change in momentum of the object, then it is given by

|\Delta p|=m|\Delta v|

- In situation 1 (pitcher throwing the ball), the change in momentum is

\Delta p = m|\Delta v|=m|v|=mv

- In situation 2 (catcher receiving the ball), the change in momentum is

\Delta p = m\Delta v = m|-v|=mv

So, the magnitude of the change in momentum is the same (but the direction is opposite)

c)

The impulse exerted on an object is equal to the change in momentum of the object:

I=\Delta p

where

I is the impulse

\Delta p is the change in momentum

As we saw in part b), the change in momentum of the ball in the two situations is the same, therefore the impulse exerted on the ball will also be the same, in magnitude.

However, the direction will be opposite, as the change in momentum has opposite direction in the two situations.

d)

To compare the time of impact in the two situations, we have to look closer into them.

- When the ball is thrown, the hand "moves together" with the ball, from back to ahead in order to give it the necessary push. We can verify therefore that the time is longer in this case.

- When the ball is cacthed, the hand remains more or less "at rest", it  doesn't move much, so the collision lasts much less than the previous situation.

Therefore, we can say that the time of impact is longer when the ball is thrown, compared to when it is catched.

e)

The impulse exerted on an object can also be rewritten as the product between the force applied on the object and the time of impact:

I=F\Delta t

where

I is the impulse

F is the force applied

\Delta t is the time of impact

This can be rewritten as

F=\frac{I}{\Delta t}

In this problem, in the two situations,

- I (the impulse) is the same in both situations

- \Delta t when the ball is thrown is larger than when it is catched

Therefore, since F is inversely proportional to \Delta t, this means that the force is larger when the ball is catched.

6 0
4 years ago
Light of wavelength 630 nm falls on two slits and produces an interference pattern in which the third-order bright red fringe is
goblinko [34]

Answer:

d \frac{x}{l} = m×  λ⇒ d = λ ×m×l / x

= 630×10^{-9} m × 3×3m/ 45×10^{-3} m

= 1.26×10^{-4}m

Explanation:

the above calculation is based on Young’s double slit experiment where the two slits provide two coherent light sources which results either constructive interference or destructive interference when passing through a double slit.

6 0
3 years ago
Determine the minimum wavelength of light absorbed by (a) diamond, (b) gallium phosphide and (c) tin sulfide if the gap energies
Talja [164]
I would say B.
I hope this helps!
7 0
3 years ago
Which of the following most logically completes the argument below?
kolezko [41]

Answer:

  (E) a greatly increased number of small particles in Earth’s orbit would result in a blanket of reflections that would make certain valuable telescope observations impossible

Explanation:

The trade is one strong reflection for many weak reflections (and more dangerous near-Earth space travel).

None of the answer choices except the last one has anything to do with the effect of exploding a satellite. When you are arguing that exploding a satellite is ill conceived, you need to address specifically the effects of exploding the satellite.

6 0
3 years ago
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