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GrogVix [38]
2 years ago
5

How long does it for a car to cover 100 miles at 60 mi/hr? Use one of the following equations:

Physics
2 answers:
olga55 [171]2 years ago
8 0

Answer:

T = 6min = 1. 10 Hour

Explanation:

x=rate of the police car

x plus 10 = rate of the sport car

Mazyrski [523]2 years ago
4 0
1 hour and 40 minutes
You might be interested in
A brick sits on the top of a hill with a gravitational potential energy of 245 J. To determine the gravitational potential of th
Bogdan [553]

Answer:

The mass of the object, its acceleration due to gravity and the distance between the top of the hill and the ground level.

Explanation:

gravitational potential energy is the energy possessed by a body under influence of gravitational force by virtue of its position.

In order to determine the gravitational potential energy of the brick, we must know the mass (m) of the brick, its acceleration due to gravity (g) since it is acting under the influence of gravitational force and the distance between the top of the hill and the ground level. (The height).

Potential energy of a body is calculated as mass × acceleration due to gravity × height.

5 0
3 years ago
8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s
allsm [11]

Answer:

a) \omega = 50\,\frac{rad}{s}, b) \omega = 0\,\frac{rad}{s}

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

\tau = F \cdot r

\tau = m\cdot a \cdot r

\tau = m \cdot \alpha \cdot r^{2}

Where \alpha is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}

\alpha = -3\,\frac{rad}{s^{2}}

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)

\omega = 50\,\frac{rad}{s}

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}

t = 66.667\,s

Since t > 66.667\,s, then the angular velocity is equal to zero. Therefore:

\omega = 0\,\frac{rad}{s}

7 0
3 years ago
Determine whether the center of mass of the system consisting of the earth and moon lies inside or outside the earth. Assume tha
tester [92]

Answer:

R_cm = 4.66 10⁶ m

Explanation:

The important concept of mass center defined by

         R_cm = 1 / M   ∑  x_i m_i

where M is the total mass, x_i and m_i are the position and masses of each body

Let's apply this expression to our case.

Let's set a reference frame where the axis points from the center of the Earth to the Moon,

       R_cm = 1 / M (m_earth 0 + m_moon d)

the total mass is

      M = m_earth + m_moon

     

the distance from the Earth is zero because all mass can be considered to be at its gravimetric center

let's calculate

      M = 5.98 10²⁴ + 7.35 10²²

      M = 6.0535 10₂⁴24 kg

we substitute

      R_cm = 1 / 6.0535 10²⁴ (0 + 7.35 10²² 3.84 )

      R_cm = 4.66 10⁶ m

4 0
3 years ago
A 3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?​
anyanavicka [17]

Answer:

I think the answer is

Explanation:

140N/m/20N =

7m is the answer

8 0
2 years ago
Read 2 more answers
Consider a river flowing toward a lake at an average speed of 3 m/s at a rate of 550 m3/s at a location 58 m above the lake surf
Vladimir [108]

Answer:

1. 0.574 kJ/kg

2. 315.7 MW

Explanation:

1. The mechanical energy per unit mass of the river is given by:

E_{m} = E_{k} + E_{p}

E_{m} = \frac{1}{2}v^{2} + gh

Where:

Ek is the kinetic energy

Ep is the potential energy

v is the speed of the river = 3 m/s

g is the gravity = 9.81 m/s²

h is the height = 58 m

E_{m} = \frac{1}{2}(3 m/s)^{2} + 9.81 m/s^{2}*58 m = 0.574 kJ/Kg

Hence, the total mechanical energy of the river is 0.574 kJ/kg.

2. The power generation potential on the river is:

P = m(t)E_{m} = \rho*V(t)*E_{m} = 1000 kg/m^{3}*550 m^{3}/s*0.574 kJ/kg = 315.7 MW

Therefore, the power generation potential of the entire river is 315.7 MW.

I hope it helps you!

4 0
3 years ago
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