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3 years ago

A stalled car is being pushed up a hill at constant velocity by three people. the net force on the car is

1 answer:

6 0

I think you need more information like the force of gravity and the force of the three people. Once you combine the two, however, you should get the net force.

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Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho

77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge $q{t}= 0.65 q_{0}$

We need to calculate the time constant

Using expression for charging in a RC circuit

$q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

Where, $\dfrac{t}{RC}$ = time constant

Put the value into the formula

$0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]$

$1-e^{-(\dfrac{t}{RC})}=0.65$

$e^{-(\dfrac{t}{RC})}=0.35$

$-\dfrac{t}{RC}=ln (0.35)$

$-\dfrac{t}{RC}=-1.049$

$\dfrac{t}{RC}=1.049$

Hence, The time constant is 1.049.

6 0

3 years ago

The combining of light nuclei is called blank. Blank as in ____. Not actually blank. You know what im tryin to say.

Andreas93 [3]

Answer:

The combining of light nuclei is called nuclear fusion.

6 0

3 years ago

Read 2 more answers

What is the acceleration of an object with a mass of 15 kg and a coefficient of friction of 0.18

11111nata11111 [884]

Answer:

a = 1.764m/s^2

Explanation:

By Newton's second law, the net force is F = ma.

The equation for friction is F(k) = F(n) * μ.

In this case, the normal force is simply F(n) = mg due to no other external forces being specified

F(n) = mg = 15kg * 9.8 m/s^2 = 147N.

F(k) = F(n) * μ = 147N * 0.18 = 26.46N.

Assuming the object is on a horizontal surface, the force due to gravity and the normal force will cancel each other out, leaving our net force as only the frictional one.

Thus, F(net) = F(k) = ma

26.46N = 15kg * a

a = 1.764m/s^2

7 0

3 years ago

The largest hailstone every measured fell in Vivian, Nebraska in 2010. The circumference of that hailstone was 19 inches. Using

Crazy boy [7]

Answer:

6.04788 in

$115.82654\ in^3$

78.38779 m/s

0.88159 kg

34.55294 J

Explanation:

Circumference is given by

$c=2\pi r\\\Rightarrow r=\dfrac{c}{2\pi}\\\Rightarrow r=\dfrac{19}{2\pi}\\\Rightarrow r=3.02394\ in$

Diameter is given by

$d=2r\\\Rightarrow d=2\times 3.02394\\\Rightarrow d=6.04788\ in$

The diameter is 6.04788 in

$6.04788\times 2.54=15.3616152\ cm$

Volume of sphere is given by

$v=\dfrac{4}{3}\pi r^3\\\Rightarrow v=\dfrac{4}{3}\pi 3.02394^3\\\Rightarrow v=115.82654\ in^3$

The volume is $115.82654\ in^3$

$115.82654\times \dfrac{1}{1728}=0.06702\ ft^3$

Fall velocity is given by

$V=k\sqrt{d}\\\Rightarrow V=20\sqrt{15.3616152}\\\Rightarrow V=78.38779\ m/s$

The velocity of the fall will be 78.38779 m/s

Mass is given by

$m=\rho v\\\Rightarrow m=29\times 0.06702\\\Rightarrow m=1.94358\ lb$

$1.94358\ lb=1.94358\times \dfrac{1}{2.20462}=0.88159\ kg$

The mass is 0.88159 kg

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2\\\Rightarrow K=\dfrac{1}{2}0.88159\times 78.38779\\\Rightarrow K=34.55294\ J$

The kinetic energy is 34.55294 J

4 0

3 years ago

What is the centripetal force that would be required to keep a 4.0 kg mass moving in a horizontal circle with a radius of 0.80 m

KIM [24]

Answer:

D. 1.8 × 102 newtons radially inward

Explanation:

The magnitude of the centripetal force is given by:

$F=m\frac{v^2}{r}$

where

m is the mass of the object

v is the tangential speed

r is the radius of the circular trajector

In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

$F=(4.0 kg)\frac{(6.0 m/s)^2}{0.80 m}=180 N$

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is

D. 1.8 × 102 newtons radially inward

4 0

4 years ago

Read 2 more answers