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2 years ago

A motorcycle accelerates from 10. m/s to 25 m/s in 5.0 seconds. What is the average acceleration of the bike?

Physics

2 answers:

I am Lyosha [343]2 years ago

5 0

Answer:

Explanation: Option (A) is the correct answer.

The data is given as follows.

Initial velocity $(v_{i})$ = 10 m/s, final velocity $(v_{f})$ = 25 m/s.

Initial time $(t_{i})$ = 0 seconds, final time $(v_{t})$ = 5 seconds.

The formula to calculate average acceleration is as follows.

Average acceleration = $\frac{v_{f} - v_{i}}{t_{f} - t_{i}}$

= $\frac{(25 - 10)m/s}{(5 - 0)s}$

= $\frac{15}{5}$

= 3 m/s/s

Thus, we can conclude that the average acceleration of the bike is 3 m/s/s.

valentinak56 [21]2 years ago

3 0

If a motorcycle accelerates from 10 m/s to 15 m/s in 5 seconds, the average acceleration of the bike is 3 m/s/s. This only means that the motorcycle’<span>s velocity will increase 3 m/s every second. You need to divide 15 m/s which is the difference of 10 and 25 to 5 seconds.</span>

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A roller coaster car drops a maximum vertical distance of 35.4 m. Determine the maximum speed of the car at the bottom of that d

marissa [1.9K]

Answer:

The maximum speed of the car at the bottom of that drop is 26.34 m/s.

Explanation:

Given that,

The maximum vertical distance covered by the roller coaster, h = 35.4 m

We need to find the maximum speed of the car at the bottom of that drop. It is a case of conservation of energy. The energy at bottom is equal to the energy at top such that :

$mgh=\dfrac{1}{2}mv^2$

$v=\sqrt{2gh}$

$v=\sqrt{2\times 9.8\times 35.4}$

v = 26.34 m/s

So, the maximum speed of the car at the bottom of that drop is 26.34 m/s. Hence, this is the required solution.

8 0

2 years ago

In the absence of air resistance, at what other angle will a thrown ball go the same distance as one thrown at an angle of 75 de

snow_tiger [21]

As we know that range of the projectile motion is given by

$R = \frac{v^2 sin(2\theta)}{g}$

here we know that range will be same for two different angles

so here we can say the two angle must be complementary angles

so the two angles must be

$\theta, 90 - \theta$

so it is given that one of the projection angle is 75 degree

so other angle for same range must be 90 - 75 = 15 degree

so other projection angle must be 15 degree

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2 years ago

A man is dragging a trunk up the loading ramp of a mover’s truck. The ramp has a slope angle of 20.0°, and the man pulls upward

AleksandrR [38]

Answer:

(a) 104 N

(b) 52 N

Explanation:

Given Data

Angle of inclination of the ramp: 20°

F makes an angle of 30° with the ramp

The component of F parallel to the ramp is Fx = 90 N.

The component of F perpendicular to the ramp is Fy.

(a)

Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.

Resolve F into its x-component from Pythagorean theorem:

Fx=Fcos30°

Solve for F:

F= Fx/cos30°

Substitute for Fx from given data:

Fx=90 N/cos30°

=104 N

(b) Resolve r into its y-component from Pythagorean theorem:

Fy = Fsin 30°

Substitute for F from part (a):

Fy = (104 N) (sin 30°)

= 52 N

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2 years ago

An green hoop with mass mh = 2.8 kg and radius rh = 0.13 m hangs from a string that goes over a blue solid disk pulley with mass

Otrada [13]

The only force on the system is the mass of the hoop F net = 2.8kg*9.81m/s^2 = 27.468 N The mass equal of the rolling sphere is found by: the sphere rotates around the contact point with the table.
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I = 7/5*mR^2 M = 7/5*m
Therefore, linear acceleration is computed by:F/m = 27.468 / (2.8 + 1/2*2 + 7/5*4) = 27.468/9.4 = 2.922 m/s^2

7 0

2 years ago

You want to move a heavy box with mass 30.0 kg across a carpeted floor. You pull hard on one of the edges of the box at an angle

charle [14.2K]

Answer:

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