A net force of 25.0 N causes an object to accelerate at 4.00 m/s2. What is the mass of the object?

Answers

Answer 1

Answer:

Answer: 7kg I think or 6

Explanation:

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1. Draw a quantitative motion map for the following description: A bicyclist speeds along a road at 10 m/s for 6 seconds. Then she stops for three seconds to make a 180˚ turn and then travels at 5 m/s for 3 seconds.

Answers

Answer:

Please find the attached file for the figure.

Explanation:

Given that a bicyclist speeds along a road at 10 m/s for 6 seconds.

Its acceleration = 10/6 = 1.667 m/s^2

The distance covered = 1/2 × 10 × 6

Distance covered = 30 m

That is, displacement = 30 m

Then she stops for three seconds to make a 180˚ turn and then travels at 5 m/s for 3 seconds.

The acceleration = 5/3 = 1.667 m/s^2

The displacement = 1/2 × 5 × 3

Displacement = 7.5 m

The resultant acceleration will be equal to zero.

While the resultant displacement will be:

Displacement = 30 - 7.5 = 22.5 m

Please find the attached file for the sketch.

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_(initial) + PE_(initial) = KE_(final) + PE_(final)$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = (1)/(2)mv^(2) + 0$

$2gH = v^(2)$

$v = √(2* 9.8* 5) = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0* 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = (1)/(2)gt^(2)$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8) = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56* 0.638 = 4.823\ m$

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

$mgh= (1)/(2)mv^2$

$v = √(2gh)$

$v = √(2* 9.8 * 5)$

v = 9.89 m/s

now he perform elastic collision

$v = (m_1v_1)/(m_1+m_2)$

$v = (65* 9.89)/(65+20)$

v = 7.57 m/s

time taken by the skies to fall is

$h = (1)/(2)gt^2$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8)}$

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m

A proton is at the origin and an electron is at the point x = 0.36 nm , y = 0.39 nm . Find the electric force on the proton.Express your answer using two significant figures. Enter your answers numerically separated by a comma.

Answers

Answer:

The electric force on the proton is 8.2x10^-10 N

Explanation:

We use the formula to calculate the distance between two points, as follows:

r = ((x2-x1)^2 + (y2-y1)^2)^1/2, where x1 and x2 are the x coordinate, y2, y1 are the y coordinate. replacing values:

r = ((0.36-0)^2 + (0.39-0)^2)^1/2 = 0.53 nm = 5.3x10^-10 m

We will use the following expression to calculate the electrostatic force:

F = (q1*q2)/(4*pi*eo*r^2)

Here we have:

q1 = q2 = 1.6x10^-19 C, 1/4*pi*eo = 9x10^9 Nm^2C^-2

Replacing values:

F = (1.6x10^-19*1.6x10^-9*9x10^9)/((5.3x10^-10)^2) = 8.2x10^-10 N

In the figure, determine the character of the collision. The masses of the blocks, and the velocities before and after are given. The collision is (Show your work-no work shown = ZERO POINTS) 1.8 m/s 0.2 m/s 0.6 m/s 1.4 m/s 4 kg 6 kg 4 kg 6 kg Before After A) perfectly elastic. B) partially inelastic. C) completely inelastic. D) characterized by an increase in kinetic energy E) not possible because momentum is not conserved.

Answers

When two bodies come into close touch with one another, a collision occurs. In this instance, the two bodies quickly exert forces on one another. The collision changes the energy and momentum of the bodies that are interacting.

Briefing

the system's initial kinetic energy, KEi, is equal to 0.5 * 4 * 1.8 2 plus 0.5 * 6 * 0.2 2 J.

KEi = 6.6 J

The system's ultimate kinetic energy, KEf

, following the collision is equal to 0.5 * 4 * 0.6 + 0.5 * 6 * 1.4 J.

KEf = 6.6 J

since KEi = KEf

Perfectly elastic is the collision

the appropriate response is A) completely elastic.

Visit: to learn more about absolutely elastic.

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