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e-lub [12.9K]
3 years ago
13

Reduce the work output

Physics
1 answer:
andriy [413]3 years ago
6 0
Maybe The 3rd One Not Sure!
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* A 5 kg ball is dropped from a height of 20 m. How much kinetic energy will it have 10 m above the ground? a. 490 J b. 392 J C.
Oksi-84 [34.3K]

Answer:

A. 490

Explanation:

soln

mass = m = 5kg

Height = h = 10m

Acceleration due to gravity = g = 9.8ms²

K.E = 1/2 × mass × (velocity)²

Recall from equations of motion

v² = u² + 2gh

Therefore,

K.E = 1/2 × mass × ( u² + 2gh)

K.E = 1/2 × 5 × ( 0² + 2×10×9.8)

K.E = 1/2 × 5 × 196

K.E = 1/2 × 980

K.E = 490 Joules

6 0
2 years ago
The kinetic theory states that the particles in matter are always in?
OlgaM077 [116]
Motion is the correct word that fits in.
Hope this helps.
7 0
3 years ago
The force of friction always opposes the _______.
Ierofanga [76]

Answer:

motion

Explanation:

uh look it up?

8 0
3 years ago
Read 2 more answers
A quarterback is set up to throw the football to a receiver who is running with a constant velocity v⃗ rv→rv_r_vec directly away
Artist 52 [7]

Answer:

a) V_o,y = 0.5*g*t_c

b) V_o,x = D/t_c - v_r

c) V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

d)  Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

Explanation:

Given:

- The velocity of quarterback before the throw = v_r

- The initial distance of receiver = r

- The final distance of receiver = D

- The time taken to catch the throw = t_c

- x(0) = y(0) = 0

Find:

a) Find V_o,y, the vertical component of the velocity of the ball when the quarterback releases it.  Express V_o,y in terms of t_c and g.

b) Find V_o,x, the initial horizontal component of velocity of the ball.   Express your answer for V_o,x in terms of D, t_c, and v_r.

c) Find the speed V_o with which the quarterback must throw the ball.  

   Answer in terms of D, t_c, v_r, and g.

d) Assuming that the quarterback throws the ball with speed V_o, find the angle Q above the horizontal at which he should throw it.

Solution:

- The vertical component of velocity V_o,y can be calculated using second kinematics equation of motion:

                               y = y(0) + V_o,y*t_c - 0.5*g*t_c^2

                              0 = 0 + V_o,y*t_c - 0.5*g*t_c^2

                               V_o,y = 0.5*g*t_c

- The horizontal component of velocity V_o,x witch which velocity is thrown can be calculated using second kinematics equation of motion:

- We know that V_i, x = V_o,x + v_r. Hence,

                               x = x(0) + V_i,x*t_c

                               D = 0 + V_i,x*t_c

                               V_o,x + v_r = D/t_c

                                V_o,x = D/t_c - v_r

- The speed with which the ball was thrown can be evaluated by finding the resultant of V_o,x and V_o,y components of velocity as follows:

                           V_o = sqrt ( V_o,x^2 + V_o,y^2)

                          V_o = sqrt ( (D/t_c - v_r)^2 + (0.5*g*t_c)^2)

       

- The angle with which it should be thrown can be evaluated by trigonometric relation:

                            tan(Q) = ( V_o,y / V_o,x )

                            tan(Q) = ( (0.5*g*t_c)/ (D/t_c - v_r) )

                                   Q = arctan ( g*t_c^2 / 2*(D - v_r*t_c) )

                           

                               

6 0
3 years ago
The barrel of a rifle has a length of 0.89 m. A bullet leaves the muzzle of a rifle with a speed of 620 m/s. What is the acceler
guapka [62]

Answer:

215955.06 m/s^2

Explanation:

length of barrel, s = 0.89 m

initial velocity of the bullet, u = 0 m/s

Final velocity of the bullet, v = 620 m/s

Let a be the acceleration of the bullet in the barrel

Use third equation of motion, we get

v^{2}=u^{2}+ 2as

620^{2}=0^{2}+ 2\times a \times 0.89

a = 215955.06 m/s^2

Thus, the acceleration of the bullet inside the barrel is  215955.06 m/s^2.

6 0
3 years ago
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