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mina [271]
2 years ago
8

An object is dropped from a bridge. A second object is thrown downwards 1.0 s later. They both reach the water 20 m below at the

same instant. What was the initial speed of the second object? Neglect air resistance.
Physics
1 answer:
Lerok [7]2 years ago
7 0
Answer:

Explanation:
Kinematics equation for first Object:

but:
The initial velocity is zero

it reach the water at in instant, t1, y(t)=0:


Kinematics equation for the second Object:
The initial velocity is zero

but:

it reach the water at in instant, t2, y(t)=0. If the second object is thrown 1s later, t2=t1-1=1.02s


The velocity is negative, because the object is thrown downwards
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5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.
andreev551 [17]

Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

a = 8/4.2

a = 1.904

6 0
2 years ago
A cart, which has a mass of 2.30 kg is sitting at the top of an inclined plane, which is 4.50 meters long and meets the horizont
expeople1 [14]

Answer:

a) The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) The magnitude of the force that causes the cart to roll down is 5.47 N.

c) The acceleration of the cart is 2.38 m/s²

d) It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

g) The work done by the gravitational force is 24.6 J.

Explanation:

Hi there!

a) The gravitational potential energy is calculated using the following equation:

EP = m · g · h

Where:

EP = gravitational potential energy.

m = mass of the object.

g = acceleration due to gravity.

h = height at which the object is located.

The height of the inclined plane can be calculated using trigonomoetry:

sin 14.0° = height / lenght

sin 14.0° = height / 4.50 m

4.50 m · sin 14.0° = height

height = 1.09 m

Then, the gravitational potential energy will be:

EP = m · g · h

EP = 2.30 kg · 9.81 m/s² · 1.09 m = 24.6 J

The gravitational potential energy before the cart rolls down the incline is 24.6 J.

b) Please, see the attached figure for a graphical description of the problem and the forces acting on the cart. The force that causes the cart to accelerate down the incline is the horizontal component of the weight (Fwx in the figure). The magnitude of this force can be obtained using trigonometry:

sin 14° = Fwx / Fw

The weight of the cart (Fw) is calculated as follows:

Fw = m · g

Fw = 2.30 kg · 9.81 m/s²

Fw = 22.6 N

Then, the x-component of the weight will be:

FW · sin 14° = Fwx

22.6 N · sin 14° = Fwx

Fwx = 5.47 N

The magnitude of the force that causes the cart to roll down is 5.47 N.

c)Using the equation of Fwx we can calculate the acceleration of the cart:

Fwx = m · a

Where "m" is the mass of the cart and "a" is the acceleration.

Fwx / m = a

5.47 N / 2.30 kg = a

a = 2.38 m/s²

The acceleration of the cart is 2.38 m/s²

d) To calculate the time it takes the cart to reach the bottom of the incline, let´s use the equation of position of the cart:

x = x0 + v0 · t + 1/2 · a · t²

Where:

x = position of the cart at time t.

x0 = initial position.

v0 = initial velocity.

a = acceleration.

t = time.

Considering the initial position as the point at which the cart starts rolling (x0 = 0) and knowing that the cart starts from rest (v0 = 0), let´s find the time it takes the cart to travel the 4.50 m of the inclined plane:

x = 1/2 · a · t²

4.50 m = 1/2 · 2.38 m/s² · t²

2 · 4.50 m / 2.38 m/s² = t²

t = 1.94 s

It takes the cart 1.94 s to reach the bottom of the incline.

e) The velocity of the cart at the bottom of the inclined plane can be obtained using the following equation:

v = v0 + a · t

v = 0 m/s + 2.38 m/s² · 1.94 s

v = 4.62 m/s

The velocity of the cart at the bottom of the inclined plane is 4.62 m/s.

f) The kinetic energy can be calculated using the following equation:

KE = 1/2 · m · v²

Where:

KE =  kinetic energy.

m = mass of the cart.

v = velocity of the cart.

KE = 1/2 · 2.30 kg · (4.62 m/s)²

KE = 24.6 J

The kinetic energy of the cart as it reaches the bottom of the incline is 24.6 J.

The gain of kinetic energy is equal to the loss of gravitational potential energy.

g) The work done by the gravitational force can be calculated using the work-energy theorem: the work done by the gravitational force is equal to the negative change in the gravitational potential energy:

W = -ΔPE

W = -(final potential energy - initial potential energy)

W = -(0 - 24.6 J)

W = 24.6 J

This can also be calculated using the definition of work:

W = Fw · d

Where "d" is the distance traveled in the direction of the force, that is the height of the inclined plane:

W = 22.6 N · 1.09 m = 24.6 J.

The work done by the gravitational force is 24.6 J.

4 0
3 years ago
A nonconducting spherical shell, with an inner radius of 4 cm and an outer radius of 6 cm, has charge spread nonuniformly throug
Vaselesa [24]

Answer:

1.57 * 10^{3} Q

Explanation:

The volume charge density is defined by ρ = \frac{Q}{V} (Equation A), where Q is the charge and V, the volume.

The units in the S.I. are \frac{Coulombs}{m^{3} }, so we have to express the radius in meters:

inner radius = 4 cm * \frac{1 m}{100 cm} = 0.04m

outer radius = 6 cm * \frac{1m}{100cm}  = 0.06m

Now, we know that the volume of the sphere is calculated by the formula:

V = \frac{4}{3}\pi r^{3}, and as we have an spherical shell, the volume is calculated by the difference between the outher and inner spheres:

V = \frac{4}{3}\pi (r_{o} ^{3} - r_{i} ^{3}), where r_{o} is the outer radius and r_{i} is the inner radius.

Replacing the volume formula in the Equation A:

ρ = \frac{Q}{\frac{4}{3}\pi(r^{3} _{o}-r_{i} ^{3})}

ρ = \frac{3Q}{4\pi (r_{o} ^{3}-r_{i} ^{3} ) }

Replacing the values of the outer and inner radius whe have:

ρ = \frac{3Q}{4\pi (1.52 * 10^{-4})}

ρ = 1.57 * 10^{3} Q

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Answer:

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