Ask question

Ask question

All categories

3 years ago

12

A sled is pulled at a constant velocity across a horizontal snow surface. If a force of 8.0 times 10^1 N is being applied to the

sled rope at an angle of 53 degrees to the grind, what is the force of friction between the sled and the snow?

1 answer:

Yanka [14]3 years ago

7 0

Answer:

48 Newtons

Explanation:

There is no acceleration and net forces.
$8*10x^{-1} N cos(53) = friction$
Answer: 48.15 Newtons

You might be interested in

What do all waves, including sound and light, have in common

Schach [20]

Answer:

They all have frequency, wavelength, amplitude, speed and also all transfer energy.

6 0

3 years ago

A 1050 kg sports car is moving westbound at 13.0 m/s on a level road when it collides with a 6320 kg truck driving east on the s

Dmitriy789 [7]

Answer:

(A) V = 9.89m/s

(B) U = -2.50m/s

(C) ΔK.E = –377047J

(D) ΔK.E = –257750J

Explanation:

The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.

This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.

See attachment below for full solution.

5 0

3 years ago

You are pushing on a heavy desk with a force of 65 N the desk does not slide the force of friction between the desk and the floo

astra-53 [7]

When an object does not move even on pushing , static frictional force acts on in opposite direction of the applied force to stop the object from moving. static frictional force is a self adjusting force and it adjust its value according to the applied force if the applied force is smaller than the maximum value of static frictional force. The object starts moving once the applied force on it becomes greater than the maximum static frictional force. hence the statement is true.

7 0

3 years ago

An object has a 600 N force pushing it to the right, while a 700 N force pushes in the opposite direction. Gravity also acts on

nordsb [41]

Left is the answer 3

6 0

3 years ago

Read 2 more answers

Establishing a potential difference The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 10

11111nata11111 [884]

Answer:

$1.62\times 10^{-8}\ \text{s}$

Explanation:

$\epsilon_0$ = Vacuum permittivity = $8.854\times 10^{-12}\ \text{F/m}$

$A$ = Area = $10\times 2\times 10^{-4}\ \text{m}^2$

$d$ = Distance between plates = 1 mm

$V_c$ = Changed voltage = 60 V

$V$ = Initial voltage = 100 V

$R$ = Resistance = $1000\ \Omega$

Capacitance is given by

$C=\dfrac{\epsilon_0A}{d}\\\Rightarrow C=\dfrac{8.854\times 10^{-12}\times 10\times 2\times 10^{-4}}{1\times 10^{-3}}\\\Rightarrow C=1.7708\times 10^{-11}\ \text{F}$

We have the relation

$V_c=V(1-e^{-\dfrac{t}{CR}})\\\Rightarrow e^{-\dfrac{t}{CR}}=1-\dfrac{V_c}{V}\\\Rightarrow -\dfrac{t}{CR}=\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-CR\ln (1-\dfrac{V_c}{V})\\\Rightarrow t=-1.7708\times 10^{-11}\times 1000\ln(1-\dfrac{60}{100})\\\Rightarrow t=1.62\times 10^{-8}\ \text{s}$

The time taken for the potential difference to reach the required level is $1.62\times 10^{-8}\ \text{s}$ .

5 0

3 years ago