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stich3 [128]
3 years ago
6

Suppose we replace the original launcher with one that fires the ball upward at twice the speed. We make no other changes. How f

ar behind the cart will the ball land, compared to the distance in the original experiment?
(A) half as far
(B)twice as far
(C) four times as far
(D) the same distance
(E) by a factor not listed above
Physics
1 answer:
expeople1 [14]3 years ago
5 0

Answer:

(C) four times as far

Explanation:

As we know that the range of the launcher is given as

R = \frac{v^2sin(2\theta)}{g}

here we know that all the parameters will remain same but only change is the velocity is doubled

So we will have

R' = \frac{(2v)^2 sin2\theta}{g}

so we have

R' = 4 R

so here correct answer will be

(C) four times as far

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A horizontal spring with spring constant 950 N/m is attached to a wall. An athlete presses against the free end of the spring, c
OLga [1]

Answer:

66.5N

Explanation:

F = kx

Where F = force

K = spring constant

x = compression

Given

K = 950N/m

x = 7.0cm

F = ?

First convert the compression to meters .

7.0cm = 7.0 x 0.01

= 0.07 meters

Therefore

F = 950 x 0.07

= 66.5N

4 0
3 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
2 years ago
Does a feather fall as fast as a rock in a vacuum? If so why?
7nadin3 [17]

Answer:

No.

Explanation:

A feather is less dense and thus less force exerted while a rock is very dense thus exerting more force .

3 0
2 years ago
Astronomers suspect that a galaxy’s type can be affected both by the conditions in the protogalactic cloud from which it forms (
Katen [24]

Answer:

The items here are describing either a condition in a later interacton or a protogalactic cloud.  The results matching with spiral and elliptical galaxy are:

For spiral galaxy are options 6,3,2 and 5.

and for elliptical galaxy are options 4 and 1.

Explanation:

Here it is given that astrnomers suspect that types of galaxy can be affected both by the conditions which occurs due to protogalactic cloud and then from it forms the initial conditions and then by the later interactions with the other galaxies.

so, both types of galaxies are matched with their respective items given:

A. Spiral galaxy:

    2. A galaxy collision results tostripping of gas.

    3. The protogalactic cloud rotates in a very slow motion.

    5. The density of protogalactic cloud is very high.

    6. when the protogalactic cloud shrinks cloud forms very rapidly.

B. Elliptical galaxy:

    1. The protogalactic cloud has high angular momentum.

    4. Most of the protogalactic gases settles down into a disk.

6 0
3 years ago
What is a non deadly thing
PIT_PIT [208]

Answer:

Air

Explanation:

4 0
3 years ago
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