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4 years ago

What is the acceleration of a 3 kg rock that is thrown with a force of 18 N?

Physics

1 answer:

Ket [755]4 years ago

7 0

Answer:6m/s^2

Explanation:

Mass =3kg

Force=18N

Acceleration =force ➗ mass

Acceleration =18 ➗ 3

Acceleration =6

Acceleration =6m/s^2

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What are the differences between refraction and diffraction?

Liula [17]

Refraction is the change in direction of a wave.
Diffraction is the bending of a wave around a barrier.

5 0

4 years ago

A metal can containing condensed mushroom soup has mass 215 g, height 10.8 cm, and diameter 6.38 cm. It is placed at rest on its

s344n2d4d5 [400]

Answer:

Part a)

Moment of inertia of the cylinder is given as

$I = 1.21 \times 10^{-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

Explanation:

As we know that the inclined plane is of length L = 3 m

and its inclination is given as 25 degree

so we know that acceleration of center of mass of the cylinder is constant so we will have

$v_f^2 = v_i^2 + 2 a L$

so we have

$v_f^2 = 0 + 2a(3)$

now we know that

$v_{avg} = \frac{L}{t} = \frac{v_f + v_i}{2}$

$\frac{3}{1.50} = \frac{v_f + 0}{2}$

$v_f = 4 m/s$

Now we have know that final speed of the cylinder due to pure rolling is given as

$v_f = \sqrt{\frac{2gH}{1 + \frac{I}{mR^2}}}$

$4 = \sqrt{\frac{2(9.81)(3 sin25)}{1 + \frac{I}{0.215(0.0319)^2}}}$

$I = 1.21 \times 10^[-4} kg m^2$

Part B)

Height of the cylinder is of no use here to calculate the inertia

Part C)

Since we don't know about the viscosity data of the soup inside the cylinder so we can't say directly about the moment of inertia of the cylinder as $I = 1/2 mR^2$

8 0

3 years ago

A coil of wire consists of 20 turns, each of which has an area of 0.0015 m2. A magnetic field is perpendicular to the surface wi

stepan [7]

Answer:

magnitude of the induced emf in the coil is 0.0153 V

Explanation:

Given data

no of turns = 20

area = 0.0015 m²

magnitude B1 = 4.91 T/s

magnitude B2 = 5.42 T/s

to find out

the magnitude of the induced emf in the coil

solution

we know here

emf = -n A d∅ /dt

so here n = 20 and

A = 0.0015

and d∅ = B2 - B1 = 5.42 - 4.91

d∅ = 0.51 T and dt at 1 sec

so put all value

emf = -n A d∅ /dt

emf = -20 (0.0015) 0.51 / 1

emf = - 0.0153

so magnitude of the induced emf in the coil is 0.0153 V

7 0

3 years ago

A solid block, with a mass of 0.15kg, on a frictionless surface is pushed directly onto a horizontal spring, with a spring const

iren [92.7K]

Answer:

16.1 m/s

Explanation:

We can solve the problem by using the law of conservation of energy.

At the beginning, the spring is compressed by x = 35 cm = 0.35 m, and it stores an elastic potential energy given by

$U=\frac{1}{2}kx^2$

where k = 316 N/m is the spring constant. Once the block is released, the spring returns to its natural length and all its elastic potential energy is converted into kinetic energy of the block (which starts moving). This kinetic energy is equal to

$K=\frac{1}{2}mv^2$

where m = 0.15 kg is the mass of the block and v is its speed.

Since the energy must be conserved, we can equate the initial elastic energy of the spring to the final kinetic energy of the block, and from the equation we obtain we can find the speed of the block:

$\frac{1}{2}kx^2=\frac{1}{2}mv^2\\v=\sqrt{\frac{kx^2}{m}}=\sqrt{\frac{(316 N/m)(0.35 m)^2}{0.15 kg}}=16.1 m/s$

4 0

4 years ago

A lever has an effort arm that is 8 meters long and the resistance arm that is 1.5 meters long, how much effort is needed to lif

ruslelena [56]

We are given with two measurements of the arm and an input weight. To answer this problem,we need to balance the forces and use the lengths of the arms.
209 N * 8 m = x * 1.5 m
x = 1114.67 N

it takes 1114.67 N to lift the input weight

6 0

3 years ago

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