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Mnenie [13.5K]
3 years ago
12

3. What is the power of a derby race car that uses a 50 N force to travel 20 meters in 10 seconds?

Physics
1 answer:
makkiz [27]3 years ago
7 0
I think it is B. 100 W
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Astronomers were at first surprised to find complicated molecules in the interstellar medium. They thought ultra-violet light fr
jeka57 [31]

Answer:

The dust present in the clouds.

Explanation:

The complicated composition molecules that can be found in space are generally associated with clouds of dust. The significant amount of dust in these clouds provides protection not only for these molecules, but for any body that makes up or is associated with dust clouds.

It is exactly this dust that protects the molecules against the action of ultraviolet rays.

8 0
2 years ago
Calculate the acceleration of a galloping horse going from 2 m/s to 12 m/s in 2 seconds.
olchik [2.2K]
A)
5m/s^2


(12m/s-2m/s)
__________ = 5m/s^2
2s
8 0
3 years ago
If Anisa swims 85.4 yards in five minutes, how many meters will she swim in 70.0 seconds? ( use the metric system for this quest
Ira Lisetskai [31]

Answer : 18.22 meters

Explanation:

1 yard. = 0.9144 meters

85.4 yards = 78.08976 meters

1 minute = 60 seconds

5 minutes = 300 seconds

Speed of Anisa = distance / time

Speed of Anisa = 78.08976 meters / 300 seconds

Speed of Anisa = 0.26029 meters / second.

Distance travelled in 70 seconds = speed * 70

Distance travelled in 70 seconds = 0.26029 * 70 = 18.22 meters

8 0
2 years ago
A truck is carrying a refrigerator as shown in the figure. The height of the refrigerator is 158.0 cm, the width is 60.0 cm. The
Arada [10]

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

<h3>What will be the maximum acceleration of the truck to avoid tipping over?</h3>

The maximum acceleration is obtained by taking clockwise moments about the tipping point of rotation.

Clockwise moment = Anticlockwise moment

Ft * 1.58 m = F * 0.67 m

where

  • Ft is tipping force = mass * acceleration, a
  • F is weight = mass * acceleration due to gravity, g

m * a * 1.58 = m * 9.81 * 0.67

a = 4.15 m/s²

The maximum acceleration the truck can have so that the refrigerator does not tip over is 4.15 m/s².

In conclusion, the acceleration of the truck is found by taking moments about the tipping point.

Learn more about moments of forces at: brainly.com/question/27282169

#SPJ1

3 0
1 year ago
A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati
Sergio [31]

Answer:

a) v = 1.01 m/s

b) a = 5.6 m/s²

Explanation:

a)

  • If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:

       \tau = I * \alpha  (1)

  • Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
  • Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
  • τ = F*r (2)
  • For a solid uniform disk, the rotational inertia regarding an axle passing through its center  is just I = m*r²/2 (3).
  • Replacing (2) and (3) in (1), we can solve for α, as follows:

       \alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)

  • Since the angular acceleration is constant, we can use the following kinematic equation:

        \omega_{f}^{2}  - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)

  • Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

       0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)

  • Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

       \omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)

  • Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

        v = \omega * r (8)

  • where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of  the disk, r becomes the radius of the disk, 0.200 m.
  • Replacing this value and (7) in (8), we get:

       v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)

b)    

  • There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

       a_{t} = \alpha * r (9)

  • where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
  • Replacing this value and (4), in (9), we get:

       a_{t}  = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)

  • Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
  • The radial acceleration is just the centripetal acceleration, that can be expressed as follows:

       a_{c} = \omega^{2} * r  (11)

  • Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
  • Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
  • Replacing this value and (7) in (11) we get:

       a_{c} = \omega^{2} * r   = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)

  • The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
  • Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:

       a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)

6 0
2 years ago
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