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saul85 [17]
1 year ago
14

Fill in the blank question.

Physics
1 answer:
puteri [66]1 year ago
4 0
Answer is on the image with the explanation. I hope that might help you with the answers

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a student pushes a 0.500 kg trolley along a frictionless surface and accelerates it from rest to 4m/s. how much kinetic energy d
aalyn [17]
  • mass=m=0.5kg
  • velocity=4m/s

\\ \sf\longmapsto KE=\dfrac{1}{2}mv^2

\\ \sf\longmapsto KE=\dfrac{1}{2}(0.5)(4)^2

\\ \sf\longmapsto KE=0.25(16)

\\ \sf\longmapsto KE=4J

7 0
2 years ago
When does the mechanical energy of a body falls freely equal twice its potential energy ?
Ilia_Sergeevich [38]
<span>As the body rises up its gravitational potential energy increases but its kinetic energy decreases.

As a body falls its gravitational potential energy decreases but it's kinetic energy increases</span>
4 0
3 years ago
A small lead ball, attached to a 1.10-m rope, is being whirled in a circle that lies in the vertical plane. The ball is whirled
hjlf

Answer:

1.84 m

Explanation:

For the small lead ball to be balanced at the tip of the vertical circle just before it is released, the reaction force , N equal the weight of the lead ball W + the centripetal force, F. This normal reaction ,N also equals the tension T in the string.

So, T = mg + mrω² = ma where m = mass of small lead ball, g = acceleration due to gravity = 9.8 m/s², r = length of rope = 1.10 m and ω = angular speed of lead ball = 3 rev/s = 3 × 2π rad/s = 6π rad/s = 18.85 rad/s and a = acceleration of normal force. So,

a = g + rω²

= 9.8 m/s² + 1.10 m × (18.85 rad/s)²

= 9.8 m/s² + 390.85 m/s²

= 400.65 m/s²

Now, using v² = u² + 2a(h₂ - h₁)  where u = initial velocity of ball = rω = 1.10 m × 18.85 rad/s = 20.74 m/s, v = final velocity of ball at maximum height = 0 m/s (since the ball is stationary at maximum height), a = acceleration of small lead ball = -400.65 m/s² (negative since it is in the downward direction of the tension), h₁ = initial position of lead ball above the ground = 1.3 m and h₂ = final position of lead ball above the ground = unknown.

v² = u² + 2a(h₂ - h₁)

So, v² - u² = 2a(h₂ - h₁)

h₂ - h₁ =  (v² - u²)/2a

h₂ =  h₁ + (v² - u²)/2a

substituting the values of the variables into the equation, we have

h₂ =  1.3 m + ((0 m/s)² - (20.74 m/s)²)/2(-400.65 m/s²)

h₂ =  1.3 m + [-430.15 (m/s)²]/-801.3 m/s²

h₂ =  1.3 m + 0.54 m

h₂ =  1.84 m

7 0
2 years ago
A straight segment of wire 35.0 cm long carrying a current of 2.60 A is in a uniform magnetic field. The segment makes an angle
den301095 [7]

Answer:

Magnetic field, B = 0.275 T

Explanation:

Given that,

Length of the wire, L = 35 cm = 0.35 m

Current carried in the wire, I = 2.6 A

The segment makes an angle of 53∘ with the direction of the magnetic field, \theta=53^{\circ}

Magnetic force, F = 0.2 N

To find,

The magnitude of the magnetic field.

Solution,

The magnetic force acting on the wire is given by :

F=ILBsin\theta

\theta is the angle between the length of wire and the magnetic field.

0.2=2.6\times 0.35\times B\times sin(53)

B = 0.275 T

Therefore, the magnitude of the magnetic field is 0.275 T. Hence, this is the required solution.

5 0
3 years ago
Question: True or False
lakkis [162]

1 True

2 True

3 True

4 False

5 False

<h3>HOPE IT HELPS.</h3>

8 0
2 years ago
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