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solniwko [45]
3 years ago
14

Samantha is refinishing her rusty wheelbarrow. She moves her sandpaper back and forth 45 times over a rusty area, each time movi

ng with a total distance of 0.12 m. Samantha pushes the sandpaper against the surface with a normal force of 2.6 N. The coefficient of friction for the metal/sandpaper interface is 0.92. How much work is done by the normal force during the sanding process
Physics
1 answer:
Leviafan [203]3 years ago
5 0

Answer:

W = 12.96 J

Explanation:

The force acting in the direction of motion of the sand paper is the frictional force. So, we first calculate the frictional force:

F = μR

where,

F = Friction Force = ?

μ = 0.92

R = Normal Force = 2.6 N

Therefore,

F = (0.92)(2.6 N)

F = 2.4 N

Now, the displacement is given as:

d = (0.12 m)(45)

d = 5.4 m

So, the work done will be:

W = F d

W = (2.4 N)(5.4 m)

<u>W = 12.96 J</u>

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liubo4ka [24]

Answer:

his results in the final angle after the collision of 37.2 degrees basically what we did there is turn the vector into a right triangle. We use sohcahtoa to solve for the angle. Being.

Explanation:

6 0
2 years ago
On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max
sesenic [268]

Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

W_{net} = 57.96 - 20 = 37.96 J

7 0
3 years ago
What does gravitational potential energy depend on?
jolli1 [7]
It depends on the mass of an object and acceleration because of the gravity and the height of an object
3 0
3 years ago
Plate Tectonic Theory
Vitek1552 [10]

Answer:

its d

Explanation:

6 0
3 years ago
The resistivity of a metal increases slightly with increased temperature. This can be expressed as rho= rho0[1+α(T−T0)] , where
Stolb23 [73]

Answer:

At 81. 52 Deg C its resistance will be 0.31 Ω.

Explanation:

The resistance of wire =R_T =\frac{\rho_T \ l}{A}

Where R_T =Resistance of wire at Temperature T

\rho_T = Resistivity at temperature T =\rho_0 \ [1 \ + \alpha\ (T-T_0\ )]

Where T_0 =20\ Deg\ C , \  \rho_0 = Constant,  \alpha =3.9 \times 10^-^3 DegC^-1 \ (Given)

l=Length of the wire

& A = Area of cross section of wire

For long and thin wire the resistance & resistivity relation will be as follows

\frac{R_T_1}{R_T_2}=\frac{\rho_0(1+\alpha \cdot(T_1-2 0 )}{\rho_0(1+\alpha \cdot (T_2 -20 )}

\frac{0.25}{0.31}=\frac{1}{[1+\alpha(T-20)]}

1.24=1+\alpha (T-20)

0.24=\alpha(\ T -20 )

Putting\ the\ value\ of \alpha = 3.9 \times 10^-^3 DegC^-1

T = 81.52 Deg C

4 0
3 years ago
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