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ANEK [815]
3 years ago
14

A 700 g can of beans is dropped from a shelf that is 1.5 m high. What is the gravitational potential energy of this can? Round y

our answer to the nearest tenth of a joule.
_____ J.
Physics
2 answers:
OLga [1]3 years ago
4 0
<span>To find the gravitational potential energy of an object, we can use this equation: GPE = mgh m is the mass of the object in kg g = 9.80 m/s^2 h is the height of the object in meters GPE = mgh GPE = (0.700 kg) (9.80 m/s^2) (1.5 m) GPE = 10.3 J The gravitational potential energy of this can is 10.3 J</span>
777dan777 [17]3 years ago
3 0

Answer:

10.3 J

Explanation:

right on edg

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Lien uses a spring scale to pull a block toward the right across the lab table. The scale reads 8 N. Which force should Lien con
lana66690 [7]

Answer:

<h2>Frictional force</h2>

Explanation:

For the block placed on the table, there are several force acting on the body along the horizontal and vertical axis. All this forces tends to keep the body in a state of equilibrium.The forces acting along the horizontal are the moving force (Fm) and the frictional force (Ff).

Frictional force are forces that acts opposite to the force that causes the body to move (moving force).

If Lien uses a spring scale to pull a block toward the right across the lab table and the scale reads 8 N, this means that the force that causes the body to move is the 8N force (moving force).

Taking the sum of force along the horizontal;

\sum fx = ma_x

Since the body is static, max = 0

\sum fx = 0\\fm+(-Ff)= 0

Note that the frictional force acts is the force of opposition acting in the negative x direction.

fm = 0+Ff\\fm = Ff

Since Fm = 8N, Ff will also be equal to 8N.

Based on the above proof, Lien can also conclude that 8 N is a frictional force

6 0
3 years ago
A 0.40 kg ball is suspended from a spring with spring constant 12 n/m . part a if the ball is pulled down 0.20 m from the equili
PolarNik [594]
The total mechanical energy of the system at any time t is the sum of the kinetic energy of motion of the ball and the elastic potential energy stored in the spring:
E=K+U= \frac{1}{2}mv^2+ \frac{1}{2}kx^2
where m is the mass of the ball, v its speed, k the spring constant and x the displacement of the spring with respect its rest position.

Since it is a harmonic motion, kinetic energy is continuously converted into elastic potential energy and vice-versa.

When the spring is at its maximum displacement, the elastic potential energy is maximum (because the displacement x is maximum) while the kinetic energy is zero (because the velocity of the ball is zero), so in this situation we have:
E=U_{max}= \frac{1}{2}k(x_{max})^2

Instead, when the spring crosses its rest position, the elastic potential energy is zero (because x=0) and therefore the kinetic energy is at maximum (and so, the ball is at its maximum speed):
E=K_{max}= \frac{1}{2}m(v_{max})^2

Since the total energy E is always conserved, the maximum elastic potential energy should be equal to the maximum kinetic energy, and so we can find the value of the maximum speed of the ball:
U_{max}=K_{max}
\frac{1}{2}k(x_{max})^2 =  \frac{1}{2}m(v_{max})^2
v_{max}= \sqrt{ \frac{k x_{max}^2}{m} }= \sqrt{ \frac{(12 N/m)(0.20 m)^2}{0.4 kg} }=1.1 m/s
3 0
3 years ago
What would be the Elastic Potential Energy (EPE) stored in a spring with a constant k = 200 N/m that is pulled to stretch 0.45m?
harkovskaia [24]

Answer:

During the summer, oxygen is supplied by surface winds. M/NPT 16 to 26 0.14 to 0.45 m 1/ 77.80 $ 28 Phone: 418 81-1 Fax: 418 81-2882 SZ5-2025 SZ22-200 SZ2-250 SZ24-00 12.5 x 60 12.5 x 60 12.5 x 60 8 x 10 10 x 15 10 x 20 15 x HI-950 can take measures between 0.1 C to 999.9 C/ F. K, J & T-type: The HI-9551

(Hope this is what you're asking for)

Hope this helps Have a good day

3 0
2 years ago
You happen to know that the coefficient of static friction between your patio table and the ground is 0.42. You decide you want
Deffense [45]

Answer:

If the force applied is larger than 185.2 N, yes.

Explanation:

In order to move the table, the pushing force must be larger than the frictional force. The frictional force is given by:

F_f = \mu mg

where

\mu=0.42 is the coefficient of static friction

m=45 kg is the mass of the table

g=9.8 m/s^2 is the gravitational acceleration

Substituting,

F_f=(0.42)(45 kg)(9.8 m/s^2)=185.2 N

So, we are able to move the table if we push with a force larger than 185.2 N.

4 0
3 years ago
If a rigid body rotates about a fixed axis passing through its center of mass, the body's linear momentum is
dexar [7]

Answer:

The linear momentum is zero

Explanation:

Because

When a rigid body is rotating about a fixed axis passing through point O, the body’s linear momentum given as L = mvG

But VG= 0 so

Linear momentum is zero

6 0
3 years ago
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