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spin [16.1K]
3 years ago
8

an object with a mass of 70kilograms is supported at a height 8meters above the ground. what's the potential energy of the objec

t with respect to the ground ?(a)560J(b)2,058J(c)5,488J(d)3,430J
Physics
1 answer:
ra1l [238]3 years ago
3 0
Gravitational Potential Energy = mgh (m=mass; g=gravitational force(9.8N/kg); h = height)

Ug = (70kg)(9.8N/kg)(8m) = 5488J which is C.)
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How long must a tow truck apply a force of 600 N to increase the speed of a 1,500 kg car at rest to 2 m/s?
slega [8]

The time the truck must apply the given force to increase its speed to given value is 5 s.

The given parameters;

  • <em>applied force, F = 600 N</em>
  • <em>mass of the truck, m = 1,500 kg</em>
  • <em>speed of the truck, v = 2 m/s</em>

The force applied to the truck is determined by Newton's second law of motion; <em>which states that the force applied to an object is directly proportional to the product of mass and acceleration of the object.</em>

F = ma

F = \frac{mv}{t} \\\\t = \frac{mv}{F} \\\\t = \frac{1500 \times 2}{600} \\\\t = 5 \ s

Thus, the time the truck must apply the given force to increase its speed to given value is 5 s.

Learn more here:brainly.com/question/1988795

7 0
3 years ago
Some help me on this please
satela [25.4K]

Answer:

Changes in the object's momentum (answer D)

Explanation:

A net force will cause an object to change its velocity, and that will affect the object's momentum, which is defined by the product of the object's mass times its velocity.

So, select the last option (D) in the given list.

8 0
3 years ago
What is the moment of inertia of a 2.0 kg, 20-cm-diameter disk for rotation about an axis (a) through the center, and (b) throug
FinnZ [79.3K]

Answer:

(a) I=0.01 kg.m²

(b) I=0.03 kg.m²

Explanation:

Given data

Mass of disk M=2.0 kg

Diameter of disk d=20 cm=0.20 m

To Find

(a) Moment of inertia through the center of disk

(b) Moment of inertia through the edge of disk

Solution

For (a) Moment of inertia through the center of disk

Using the equation of moment  of Inertia

I=\frac{1}{2}MR^{2}\\  I=\frac{1}{2}(2.0kg)(0.20m/2)^{2}\\  I=0.01 kg m^{2}

For (b) Moment of inertia through the edge of disk

We can apply parallel axis theorem for calculating moment of inertia

I=(1/2)MR^{2}+MD\\ Here\\D=R\\I=(1/2)(2.0kg)(0.20m/2)^{2}+(2.0kg)(0.20m/2)^{2}\\  I=0.03kgm^{2}

8 0
3 years ago
2. A construction consultant may be responsible for
Brrunno [24]

This is a tricky one but on my part I'd have to say depending on the contract A,B,C.

3 0
4 years ago
How should the student change the circuit to give negative values for current and
dmitriy555 [2]

Answer:

Flip the cell.

Explanation:

This reverses direction of energy transfer.

Alternatively, flip ammeters and voltmeters to give negative readings.

what do penguins eat for lunch?

Ice-burgers!

3 0
3 years ago
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