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3 years ago

PLEASE ANSWER ASAP

Physics

1 answer:

kirill [66]3 years ago

4 0

Answer:

55.44.4 hjejjejjejej be

Explanation:

hehhehebebebbebeb

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What is the relationship between mass and weight in physics?

Alex_Xolod [135]

We are aware that weight is the product of applied gravitational force and mass. W = MG thus, where W represents the weight, M the mass, and G the gravitational force. As a result, it might also mean that "an object's weight is directly proportionate to its mass."

Mass is a physical body's total amount of matter.
It also serves as a gauge for the body's inertia, or resistance to acceleration (change in velocity) in the presence of a net force.
The strength of an object's gravitational pull to other bodies is also influenced by its mass.
The kilogram is the primary mass unit in the SI (kg).
Even though weight is frequently measured using a spring scale rather than a balancing scale and directly compared with known masses, mass is not the same as weight in physics.

<h3>What is weight?</h3>

The force exerted on an object by gravity is known as the weight of the object in science and engineering.
Weight is sometimes described as a vector quantity, or the gravitational force exerted on the object, in some common textbooks.
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2 years ago

A 10 kg ball is held above a building with a height of 30 m. What is the

Darina [25.2K]

Answer: 2940 J

Explanation: solution attached:

PE= mgh

Substitute the values:

PE= 10kg x 9.8 m/s² x 30 m

= 2940 J

6 0

4 years ago

Why does a third class lever cannot magnify force?

maw [93]

Explanation:

The third class lever cannot magnify our force because in third class lever the effort it between the load and the fulcrum. Also, in this type of lever no matter where the force is applied, it is always greater than the force of load. Hence, That type of lever cannot magnify our force.

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3 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

1. A small package is dropped from the Golden Gate Bridge. What is the velocity of the package

Sidana [21]

The velocity of the package after it has fallen for 3.0 s is 29.4 m/s

From the question,

A small package is dropped from the Golden Gate Bridge.

This means the initial velocity of the package is 0 m/s.

We are to calculate the velocity of the package after it has fallen for 3.0 s.

From one of the equations of kinematics for objects falling freely,

We have that,

v = u + gt

Where

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

and t is time

To calculate the velocity of the package after it has fallen for 3.0 s

That means, we will determine the value of v, at time t = 3.0 s

The parameters are

u = 0 m/s

g = 9.8 m/s²

t = 3.0 s

Putting these values into the equation

v = u + gt

We get

v = 0 + (9.8×3.0)

v = 0 + 29.4

v = 29.4 m/s

Hence, the velocity of the package after it has fallen for 3.0 s is 29.4 m/s

Learn more here: brainly.com/question/13327816

6 0

3 years ago