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2 years ago

7

When a car goes around a circular curve on a horizontal road at constant speed, what force causes it to follow the circular path

?

1 answer:

Reil [10]2 years ago

4 0

Centripetal force allows cars to follow a circular path.

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A packing crate with mass 80.0 kg is at rest on a horizontal, frictionless surface. At t = 0 a net horizontal force in the +x-di

Nataly [62]

Answer:

Final speed of the crate is 15 m/s

Explanation:

As we know that constant force F = 80 N is applied on the object for t = 12 s

Now we can use definition of force to find the speed after t = 12 s

$F . t = m(v_f - v_i)$

so here we know that object is at rest initially so we have

$80 (12) = 80( v_f - 0)$

$v_f = 12 m/s$

Now for next 6 s the force decreases to ZERO linearly

so we can write the force equation as

$F = 80 - \frac{40}{3} t$

now again by same equation we have

$\int F .dt = m(v_f - v_i)$

$\int (80 - (40/3)t) dt = 80(v_f - 12)$

$80 t - \frac{40t^2}{6} = 80(v_f - 12)$

put t = 6 s

$480 - 240 = 80(v_f - 12)$

$v_f = 12 + 3$

$v_f = 15 m/s$

6 0

2 years ago

The image shows positions of the earth and the moon in which region would an astronaut feel the lightest

trapecia [35]

Answer:

The moon region

Explanation:

This is because there is little to no gravity on the moon. That is where the astronaut would feel the lightest.

5 0

2 years ago

Read 2 more answers

The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

SashulF [63]

Answer:

7.9 $\frac{m}{s^{2} }$

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

$\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\$

Using algebra, we can rearrange our equation to find the final acceleration, $a_{2}$ :

$a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\$

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 $\frac{m }{s^{2} }$ = 8.33 $\frac{m }{s^{2} }$

Plug everything in:

7.9 $\frac{m }{s^{2} }$ = $\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}$

(1590kg the initial weight plus the weight of the added passenger)

8 0

1 year ago

A 2.5-L tank initially is empty, and we want to fill it with 10 g of ammonia. The ammonia comes from a line with saturated vapor

Alex17521 [72]

Answer:

592.92 x 10³ Pa

Explanation:

Mole of ammonia required = 10 g / 17 =0 .588 moles

We shall have to find pressure of .588 moles of ammonia at 30 degree having volume of 2.5 x 10⁻³ m³. We can calculate it as follows .

From the relation

PV = nRT

P x 2.5 x 10⁻³ = .588 x 8.32 x ( 273 + 30 )

P = 592.92 x 10³ Pa

3 0

2 years ago

Two ice skaters, Lilly and John, face each other while stationary and push against each others hands. John's mass is twice the m

VladimirAG [237]

Answer:

lily's speed would be twice john's speed

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2 years ago