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WINSTONCH [101]
2 years ago
11

. Prior to Remy's trip to Cleveland, his uncle tells him about this amazing barbecue restaurant there and raves about

Physics
1 answer:
Andrews [41]2 years ago
6 0

whats the restraunt called

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You are a support technician working in a data closet in a remote office. You suspect that a connectivity problem is related to
Nostrana [21]

Answer:

crimping tool

Explanation:

This is a tool employed in affixing a connector to the end of a network cable.

5 0
3 years ago
Mr. Phillips' car is parked on a steep hill with the brakes applied and the engine off. Because of the car's position, it has gr
aivan3 [116]

Answer:

The anser is b

Explanation:

4 0
3 years ago
An elevator car weighs 5500 N. If the car accelerates upwards at a rate of 4.0 m/s2, what is the tension in the support cable li
nadya68 [22]

Answer:

Explanation:

The equation for this, since we are talking about weight on an elevator, is Newton's 2nd Law adjusted to fit our needs:

F_n=ma+w where the Normal Force needed to lift that elevator car is the tension. So the equation then becomes

T = ma + w where T is the tension in the cable to lift the elevator, m is the mass of the elevator (which we have to solve for), a is the acceleration of the elevator (positive since it's going up), and w is the weight of the elevator (which we have as 5500 N). Solving first for mass:

w = mg and

5500 =- m(10) so

m = 550 kg. Now we have what we need to solve for the tension:

T = 550(4.0) + 5500 and

T = 2200 + 5500 so

T = 7700 N

4 0
3 years ago
If the coefficient of static friction between your coffee cup and the horizontal dashboard of your car is µs = 0.800, how fast c
Cloud [144]

Answer:

Before start of slide velocity will be 14.81 m/sec

Explanation:

We have given coefficient of static friction \mu =0.8

Angle of inclination is equal to \Theta =tan^{-1}\mu

\Theta =tan^{-1}0.8=38.65^{\circ}

tan{38.65^{\circ}}=0.8

Radius is given r = 28 m

Acceleration due to gravity g=9.8m/sec^2

We know that tan\Theta =\frac{v^2}{rg}

0.8=\frac{v^2}{28\times 9.8}

v^2=219.52

v=14.816m/sec

So before start of slide velocity will be 14.81 m/sec

3 0
3 years ago
Work-Energy Theorem: A 4.0 kg object is moving with speed 2.0 m/s. A 1.0 kg object is moving with speed 4.0 m/s. Both objects en
allsm [11]

The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:

FΔx = 0.5mv²

F = force, Δx = distance traveled, m = mass, v = speed

Isolate Δx:

Δx = 0.5mv²/F

Calculate Δx for each object.

Object 1: m = 4.0kg, v = 2.0m/s

Δx = 0.5(4.0)(2.0)²/F = 8/F

Object 2: m = 1.0kg, v = 4.0m/s

Δx = 0.5(1.0)(4.0)²/F = 8/F

The two objects travel the same distance before stopping.

4 0
3 years ago
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