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joja [24]
3 years ago
9

A 50 g dart rests up against a spring that has been compressed 0.04 meters. It has a spring constant of 1560 N/m.What is the max

imum velocity of the dart after the spring has transferred all its energy to it.
Physics
1 answer:
nataly862011 [7]3 years ago
5 0

Answer:

The maximum velocity of the dart after the spring has transferred all its energy to it is 7.06 m/s.  

Explanation:

Given that,

Mass of the dart, m = 50 g = 0.05 kg

Compression in the spring, x = 0.04 m

Spring constant, k = 1560 N/m

We need to find the maximum velocity of the dart after the spring has transferred all its energy to it. The kinetic energy is converted to elastic potential energy as :

\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2

v=\sqrt{\dfrac{kx^2}{m}} \\\\v=\sqrt{\dfrac{1560\times (0.04)^2}{0.05}} \\\\v=7.06\ m/s

So, the maximum velocity of the dart after the spring has transferred all its energy to it is 7.06 m/s.

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SOMEBODY PLEASE HELP!!! Indicate the reasons why the centripetal acceleration (and centripetal force) always point to the center
maxonik [38]
Well we know the correct answer cannot be "a" bcause velocity is tangent to the circlular path of an object experienting centripical motion. Velocity DOES NOT point inward in centripical motion.

we know the correct answer cannot be "b" because "t" stands for "time" which cannot point in any direction. so, time cannot point toward the center of a circle and therefore this answer must be incorrect.

I would choose answer choice "c" because both force and centripical acceleration point toward the center of the circle.

I do not think answer choice "d" can be correct because the velocity of the mass moves tangent to the circle. velocity = (change in position) / time. Therefore, by definition the mass is moving in the direction of the velocity which does not point to the center of the circle.

does this make sense? any questions?

3 0
2 years ago
It is estimated that 26 large pizzas are about right to serve 66 students of a physics club meeting. How many pizzas would be re
Brums [2.3K]

Answer:

The answer to your question is: 15 pizzas

Explanation:

data

26 large pizzas ------ 66 students

? large pizzas --------   38 students

Rule of three

x = 38 (26) / 66 = 14.96 ≈ 15 pizzas

5 0
3 years ago
A box falls off of a tailgate and slides along the street for a distance of 62.5 m. Friction slows the box at –5.0 m/s2. At what
Mila [183]

Answer:

25 m/s

Explanation:

This question can be solved using equation of motion

v^2 = u^2 + 2as

where

v is the final velocity

u is the initial velocity

s is the distance covered while moving from initial to final velocity

a is the acceleration

_____________________________________________

Given

box moved for distance of 62.5 m

Friction slows the box at –5.0 m/s2----> this statement means that there is deceleration , speed of truck decreases by 5 m/s in every second until the box comes to rest. Friction causes this deceleration.

thus in this problem

a = -5.0 m/s2

V = 0   as body came to rest due to friction deceleration

u the initial velocity we have to find

the initial velocity of box will be the same as speed of truck, as the box was in the truck and hence box will pick the speed of truck.

so if we find speed of box, we will be able get sped of truck as well.

using equation of motion

v^2 = u^2 + 2as\\0^2 = u^2 + 2*-5* 62.5\\0 = u^2 - 625\\u^2 = 625\\\sqrt{u^2} = \sqrt{625} \\u = 25

Thus, initial speed with the truck was travelling was 25 m/s.

3 0
2 years ago
A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of
Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

h=ut-\dfrac{1}{2}gt^2

h=(u\sin60)\times t-\dfrac{1}{2}gt^2

Put the value into the formula

h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2

h=41.6\ m

(b). We need to calculate the maximum height of the ball

Using formula of height

h_{max}=\dfrac{(u\sin\theta)^2}{2g}

Put the value into the formula

h=\dfrac{(33\sin60)^2}{2\times 9.8}

h=41.67\ m

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

v=u-gt

v=u\sin\theta-gt

Put the value into the formula

v_{y}=33\times\sin 60-9.8\times3.0

v_{y}=-0.82\ m/s

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

v_{x}=u\cos\theta

Put the value into the formula

v_{x}=33\times\cos60

v_{x}=16.5\ m/s

We need to calculate the ball's impact speed

Using formula of velocity

v=\sqrt{v_{x}^2+v_{y}^2}

Put the value into the formula

v=\sqrt{(16.5)^2+(-0.82)^2}

v=16.52\ m/s

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0
2 years ago
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