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Marianna [84]
2 years ago
12

This is a two part question: What is the force of friction acting on a 100 kg steel slab at rest on a steel floor?

Physics
1 answer:
Crazy boy [7]2 years ago
7 0
1) 0N... friction opposes the motion of an object, since the block is at rest there is no motion thus no friction

2) F=ma
= (5.5kg)(30m/s)
=165 N
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During the annual shuffleboard competition, Renee gives her puck an initial speed of 8.12 m/s. Once leaving her stick, the puck
shutvik [7]
<h2>Answer:2.65 seconds</h2>

Explanation:

Let a be the acceleration.

Let u be the initial velocity.

Let v be the final velocity.

Let t be the time taken.

As we know from the equations of motion,

v=u+at

Given,v=0\\u=8.12ms^{-1}\\a=-3.06ms^{-2}

0=8.12-3.06t

t=2.65sec

6 0
3 years ago
SOH-CAH-TOA is used to solve for the ________
Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

3 0
3 years ago
A coaxial cable consists of a solid inner cylindrical conductor of radius 2 mm and an outer cylindrical shell of inner radius 3
4vir4ik [10]

Answer:

d) 1.2 mT

Explanation:

Here we want to find the magnitude of the magnetic field at a distance of 2.5 mm from the axis of the coaxial cable.

First of all, we observe that:

- The internal cylindrical conductor of radius 2 mm can be treated as a conductive wire placed at the axis of the cable, since here we are analyzing the field outside the radius of the conductor. The current flowing in this conductor is

I = 15 A

- The external conductor, of radius between 3 mm and 3.5 mm, does not contribute to the field at r = 2.5 mm, since 2.5 mm is situated before the inner shell of the conductor (at 3 mm).

Therefore, the net magnetic field is just given by the internal conductor. The magnetic field produced by a wire is given by

B=\frac{\mu_0 I}{2\pi r}

where

\mu_0 is the vacuum permeability

I = 15 A is the current in the conductor

r = 2.5 mm = 0.0025 m is the distance from the axis at which we want to calculate the field

Substituting, we find:

B=\frac{(4\pi\cdot 10^{-7})(15)}{2\pi(0.0025)}=1.2\cdot 10^{-3}T = 1.2 mT

8 0
3 years ago
A box weighing 12,000 N is parked on a 36° slope. What will be the component of the weight parallel to the plane that balances f
sweet-ann [11.9K]

Given :

A box weighing 12,000 N is parked on a 36° slope.

To Find :

What will be the component of the weight parallel to the plane that balances friction.

Solution :

The component of that will be parallel to the plane to balance friction is :

F \ cos ( 90 - 36)^o\\\\F\ ( sin\ 36^o)

Therefore, component of force to balance friction is F sin 36° .

Hence, this is the required solution.

5 0
2 years ago
For 0.37 moles of oxygen (02) gas at room temperature with active translational and rotational degrees of freedom.
Lelechka [254]

Answer:

B. 161.5 J

Explanation:

n = Number of moles = 0.37

\Delta T = Rise in the temperature of the oxygen gas = 15 K

Q = heat added in order to raise the temperature

c_{p} = specific heat at constant pressure = 3.5

At constant pressure, heat is given as

Q = n c_{p} R \Delta T\\Q = (3.5) (0.37) (8.314) (15)\\Q = 161.5 J

6 0
3 years ago
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