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marta [7]
3 years ago
14

A car traveled a distance of 80 km in 45 minutes (0.75 hours). What was the speed of the car? A. 60 km/hr B. 35 km/hr C. 26 km/h

r D. 107 km/hr
Physics
1 answer:
pshichka [43]3 years ago
8 0

Answer:

The answer is 107 km/hr

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Clouds in our atmosphere cause us to see the different phases of the moon.
krok68 [10]

Answer:

False

Explanation:

:)

8 0
4 years ago
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Even when the head is held erect, as in the figure below, its center of mass is not directly over the principal point of support
alexandr1967 [171]

We are asked to determine the force required by the neck muscle in order to keep the head in equilibrium. To do that we will add the torques produced by the muscle force and the weight of the head. We will use torque in the clockwise direction to be negative, therefore, we have:

\Sigma T=r_{M\perp}(F_M)-r_{W\perp}(W)

Since we want to determine the forces when the system is at equilibrium this means that the total sum of torque is zero:

r_{M\perp}(F_M)-r_{W\perp}(W)=0

Now, we solve for the force of the muscle. First, we add the torque of the weight to both sides:

r_{M\perp}(F_M)=r_{W\perp}(W)

Now, we divide by the distance of the muscle:

(F_M)=\frac{r_{W\perp}(W)}{r_{M\perp}}

Now, we substitute the values:

F_M=\frac{(2.4cm)(50N)}{5.1cm}

Now, we solve the operations:

F_M=23.53N

Therefore, the force exerted by the muscles is 23.53 Newtons.

Part B. To determine the force on the pivot we will add the forces we add the vertical forces:

\Sigma F_v=F_j-F_M-W

Since there is no vertical movement the sum of vertical forces is zero:

F_j-F_M-W=0

Now, we add the force of the muscle and the weight to both sides to solve for the force on the pivot:

F_j=F_M+W

Now, we plug in the values:

F_j=23.53N+50N

Solving the operations:

F_j=73.53N

Therefore, the force is 73.53 Newtons.

8 0
1 year ago
A 2.00-kg object A is connected with a massless string across a massless, frictionless pulley to a 3.00-kg object B. Object A re
slamgirl [31]

Answer:

  • tension: 19.3 N
  • acceleration: 3.36 m/s^2

Explanation:

<u>Given</u>

  mass A = 2.0 kg

  mass B = 3.0 kg

  θ = 40°

<u>Find</u>

  The tension in the string

  The acceleration of the masses

<u>Solution</u>

Mass A is being pulled down the inclined plane by a force due to gravity of ...

  F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N

Mass B is being pulled downward by gravity with a force of ...

  F = mg = (3 kg)(9.8 m/s^2) = 29.4 N

The tension in the string, T, is such that the net force on each mass results in the same acceleration:

  F/m = a = F/m

  (T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)

  T = (2(29.4) +3(12.5986))/5 = 19.3192 N

__

Then the acceleration of B is ...

  a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2

The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.

3 0
4 years ago
An example of a Natural electric field is
liberstina [14]

Answer:

Lightning

Explanation

6 0
4 years ago
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Momentum for a system can be conserved in one direction while not being conserved in another. What is the anglebetween the direc
katrin [286]

Answer:

90°

Explanation:

The angle will be 90° when momentum for a system can be conserved in one direction while not being conserved in another.

The example can be

If we apply force on an object horizontally in west direction, then as in other direction south or north we cannot apply the principal of momentum conservation.

5 0
3 years ago
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