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JulijaS [17]
3 years ago
7

When parking uphill on a two-way street with no curb, your front door wheels should be

Physics
1 answer:
liubo4ka [24]3 years ago
7 0

When parking uphill on a two-way street with no curb, your front wheels should be: Turned to the right (toward the street). When you come to a corner where there are no traffic lights or stop signs, you should: Slow down, so you can stop, if necessary.

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A pycnocline is a boundary between changing regions of which of the following properties? What about a thermocline?
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5 0
3 years ago
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A man supports himself and the uniform horizontal beam pulling the rope with a force T.The weights of men and the beam are 883 N
artcher [175]

Answer:

T=502.5N

Ax=171.8N

Explanation:

The computation of the tension T in the rope and the forces exerted by the pin at A is shown below:

vertical forces sum = Ay + Tsin20 + T - 245 - 883 = 0

Now  

horizontal forces sum = Ax - Tcos70

Now Moment about B

-Ay × 4.8 + 245 × 2.4 + 883 × 1.8=0

Ay=453.6N

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Ax=171.8N

3 0
2 years ago
A clarinetist, setting out for a performance, grabs his 3.070 kg clarinet case (including the clarinet) from the top of the pian
Cerrena [4.2K]

Answer:

the vertical acceleration of the case is 1.46 m/s

Explanation:

Given;

mass of the clarinet case, m = 3.07 kg

upward force applied by the man, F = 25.60 N

Apply Newton's second law of motion;

the upward force on the clarinet case = its weight acting downwards + downward force due to its downward accelaration

F = mg + m(-a)

the acceleration is negative due to downward motion from the top of the piano.

F = mg - ma

ma = mg - F

a = \frac{mg - F}{m} \\\\a = \frac{(3.07 \times 9.8) \ - \ 25.6}{3.07} \\\\a = 1.46 \ m/s^2

Therefore, the vertical acceleration of the case is 1.46 m/s²

4 0
3 years ago
A rope pulls a 82.5 kg skier at a constant speed up a 18.7° slope with μk = 0.150. How much force does the rope exert?
Artist 52 [7]

Answer:

374 N

Explanation:

N = normal force acting on the skier

m = mass of the skier = 82.5

From the force diagram, force equation perpendicular to the slope is given as

N = mg Cos18.7

μ = Coefficient of friction = 0.150

frictional force is given as

f = μN

f =  μmg Cos18.7

F = force applied by the rope

Force equation parallel to the slope is given as

F - f - mg Sin18.7 = 0

F - μmg Cos18.7 - mg Sin18.7 = 0

F = μmg Cos18.7 + mg Sin18.7

F = (0.150 x 82.5 x 9.8) Cos18.7 + (82.5 x 9.8) Sin18.7

F = 374 N

6 0
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