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MAXImum [283]
3 years ago
14

Peter designed a road with a curve of radius 30 m that is banked so that a 950 kg car traveling at 40.0 km/h can round it even i

f the road is so icy that the coefficient of static friction is approximately zero. Neglect the effect of air drug and rolling friction. If the coefficient of static friction between and the road and the tires is .30 what is the safe range of speed?
Physics
1 answer:
spayn [35]3 years ago
3 0

Answer:

v = 15.56 m/s

v = 56 km/h

Explanation:

When coefficient of friction is approximately zero then we have

F_ncos\theta = mg

F_n sin\theta = \frac{mv^2}{R}

tan\theta = \frac{v^2}{Rg}

here we know that

v = 40 km/h = 11.11 m/s

R = 30 m

tan\theta = \frac{11.11^2}{30\times 9.81}

\theta = 22.75 degree

now when friction coefficient is 0.30 then we have

F_n cos\theta = mg + F_f sin\theta

F_f cos\theta + F_n sin\theta = \frac{mv^2}{R}

now we have

v = \sqrt{Rg(\frac{\mu + tan\theta}{1 - \mu tan\theta})}

v = \sqrt{30(9.81)(\frac{0.30 + tan22.75}{1 - (0.30) tan22.75})}

v = 15.56 m/s

v = 56 km/h

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A 125-kg astronaut (including space suit) acquires a speed of 2.50 m/s by pushing off with her legs from a 1900-kg space capsule
ryzh [129]

(a) 0.165 m/s

The total initial momentum of the astronaut+capsule system is zero (assuming they are both at rest, if we use the reference frame of the capsule):

p_i = 0

The final total momentum is instead:

p_f = m_a v_a + m_c v_c

where

m_a = 125 kg is the mass of the astronaut

v_a = 2.50 m/s is the velocity of the astronaut

m_c = 1900 kg is the mass of the capsule

v_c is the velocity of the capsule

Since the total momentum must be conserved, we have

p_i = p_f = 0

so

m_a v_a + m_c v_c=0

Solving the equation for v_c, we find

v_c = - \frac{m_a v_a}{m_c}=-\frac{(125 kg)(2.50 m/s)}{1900 kg}=-0.165 m/s

(negative direction means opposite to the astronaut)

So, the change in speed of the capsule is 0.165 m/s.

(b) 520.8 N

We can calculate the average force exerted by the capsule on the man by using the impulse theorem, which states that the product between the average force and the time of the collision is equal to the change in momentum of the astronaut:

F \Delta t = \Delta p

The change in momentum of the astronaut is

\Delta p= m\Delta v = (125 kg)(2.50 m/s)=312.5 kg m/s

And the duration of the push is

\Delta t = 0.600 s

So re-arranging the equation we find the average force exerted by the capsule on the astronaut:

F=\frac{\Delta p}{\Delta t}=\frac{312.5 kg m/s}{0.600 s}=520.8 N

And according to Newton's third law, the astronaut exerts an equal and opposite force on the capsule.

(c) 25.9 J, 390.6 J

The kinetic energy of an object is given by:

K=\frac{1}{2}mv^2

where

m is the mass

v is the speed

For the astronaut, m = 125 kg and v = 2.50 m/s, so its kinetic energy is

K=\frac{1}{2}(125 kg)(2.50 m/s)^2=390.6 J

For the capsule, m = 1900 kg and v = 0.165 m/s, so its kinetic energy is

K=\frac{1}{2}(1900 kg)(0.165 m/s)^2=25.9 J

3 0
3 years ago
Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

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3 years ago
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