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Vlad1618 [11]
3 years ago
9

How high does Pete lift his sledge hammer if he used a force of 25N to lift the hammer while doing 50J of work?

Physics
1 answer:
sasho [114]3 years ago
8 0

Answer:He lifts 2 meters

Explanation:We are trying to find the distance. The formula for distance is W/Force. 50 is our amount of work. 25N is how much force was used. Divide the work bye the force. 50/25=2M

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A car with a mass of 1380 Kg is traveling at 23 m/s to the north. A truck with a mass of 1625 Kg is traveling at 26 m/s to the s
trasher [3.6K]

Answer: -3.49 m/s (to the south)

Explanation:

This problem can be solved by the Conservation of Momentum principle which establishes the initial momentum p_{i} must be equal to the final momentum p_{f}, and taking into account this is aninelastic collision:

Before the collision:

p_{i}=mV_{o}+MU_{o} (1)

After the collision:

p_{f}=(m+M)V_{f} (2)

Where:

m=1380 kg is the mass of the car

V_{o}=23 m/s is the velocity of the car, directed to the north

M=1625 kg is the mass of the truck

U_{o}=-26 m/s is the velocity of the truck, directed to the south

V_{f} is the final velocity of both the car and the truck

p_{i}=p_{f} (3)

mV_{o}+MU_{o}=(m+M)V_{f} (4)

Isolating V_{f}:

V_{f}=\frac{mV_{o}+MU_{o}}{m+M} (5)

V_{f}=\frac{(1380 kg)(23 m/s)+(1625 kg)(-26 m/s)}{1380 kg+1625 kg} (6)

Finally:

V_{f}=-3.49 m/s The negative sign indicates the direction of the velocity is to the south

8 0
3 years ago
Line segment gj is a diameter of circle l. angle k measures (4x 6)°. circle l is inscribed with triangle g j k. line segment g j
choli [55]

The value of x in the given right triangle in a semicircle is determined as 21.

<h3>What is the measure of a triangle in a semicircle?</h3>

The triangle in a semicircle is always a right angle triangle.

From the figure shown, we can say that the triangle  G J K is right triangle and m<K = 90degrees.

Given that m<K = 4x + 6, we will can use the following equation to find the value of x as shown:

4x + 6 = 90

4x = 90 - 6

4x = 84

x = 21

Thus, the value of x in the given right triangle in a semicircle is determined as 21.

Learn more about right angle here: brainly.com/question/64787

#SPJ4

8 0
2 years ago
How can humidity be used to predict if it is going to rain?
postnew [5]
1. Humidity cannot be used to predict rain.
2. I'm pretty sure it's weather but I'm not 100% sure. Maybe like 89% sure.

3. Tempurature doesn't affect humidity.

4. Not sure but I think its the 3rd one
3 0
3 years ago
The wave length of violet light rounded to the nearest nanometer is a __ nm
nataly862011 [7]
Wavelength= speed / frequency 

so.....3× 10^8 / 7.26×10^14
= .413× 10^(-6)

in scientific notation= 4.13×10^(-7) 

in nanometer = 413 nm
6 0
3 years ago
Jake is helping Fin push a box at a constant velocity up an incline that makes an angle of 30.0° above the horizontal by applyin
andre [41]

Given data

The angle of inclination of the plane is theta = 30 degree

The applied force in the inclined plane is F = 94 N

The distance moved in the inclined plane is d = 2.30 m

The coefficient of kinetic friction is u_k = 0.280

The free-body diagram of the above configuration is shown below:

Here, the normal reaction force on the box is N, the acceleration due to gravity is denoted as g, the friction force on the box is F_f, and the mass of the box is denoted as m.

(a)

The expression for the work done by the pushing force is given as:

W=Fd

Substitute the value in the above equation.

\begin{gathered} W=94\text{ N}\times2.30\text{ m} \\ W=216.2\text{ J} \end{gathered}

Thus, the work done by the pushing force is 216.2 J.

(b)

The box is moving at the constant velocity, therefore, the pushing force will be equal to the frictional force and the component of the gravitational force in the inclined plane.

\begin{gathered} F=F_f+mg\sin \theta \\ F=\mu_kN+mg\sin \theta \end{gathered}

The expression for the normal reaction force is given as:

N=mg\cos \theta

The expression for the mass of the box is given as:

\begin{gathered} F=\mu_k\times mg\cos \theta+mg\sin \theta \\ m=\frac{F}{\mu_kg\cos \theta+g\sin \theta} \end{gathered}

Substitute the value in the above equation.

\begin{gathered} m=\frac{94\text{ N}}{0.28\times9.8m/s^2\times\cos 30^o+9.8m/s^2\times\sin 30^0} \\ m=12.9\text{ kg} \end{gathered}

Thus, the mass of the box is 12.9 kg.

7 0
1 year ago
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