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leonid [27]
3 years ago
14

You are riding a bicycle. If you apply a forward force of 150 N, and you and the bicycle have a combined mass of 90 kg, what wil

l be the forward acceleration of the bicycle Assume there is no friction)
A. 1.67 m/s2
B. 1.85 m/s2 m
C. 0.60 m/s2
D. 3.37 m/s2​
Physics
1 answer:
bekas [8.4K]3 years ago
6 0
<h3>Hello There!!</h3>

<h3><u>Given</u>,</h3>

Force(F) = 150N

Mass(m) = 90kg

<h3><u>To </u><u>Find,</u></h3>

Acceleration(a) = ?

<h3><u>We know,</u></h3>

F= m×a

150 = 90 \times  \text{a} \\  \\  \text{a} =  \frac{150}{90}  \\  \fbox{cancelling by 3} \\  \\   \text{ a}  = \cancel \frac{150}{90} \\  \\ \text{ a}  =  \frac{5}{3}  = 1.67 \text{m/s} {}^{2}

\therefore  \text{Option A= 1.67 m/s² is the correct answer}

<h3>Hope this helps</h3>
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yawa3891 [41]

The work done by the man pushing the car over the given distance is 1000J.

Given the data in the question;

  • Mass of car; m = 200kg
  • Acceleration of the car; a = 0.5m/s^2
  • Distance covered by the car; d = 10m

Work done; W = \ ?

<h3>Work done</h3>

Work done is simply defined as the energy transfer that takes place when an object is either pushed or pulled over a certain distance by an external force. It is expressed as;

Work\ done = f * d

Where f is force applied and d is distance travelled.

To determine the work done by the man, we first solve for the force applied F.

From Newton's Second Law; Force \ F = m * a

We substitute our given values into the expression

F = m * a \\\\F = 200kg * 0.5m/s^2\\\\F = 100kg.m/s^2

Next we substitute our values into the expression of work done above.

Work \ done = f * d\\\\Work \ done = 100kg.m/s^2 * 10m\\\\Work \ done = 1000kgm^2/s^2\\\\Work \ done = 1000J

Therefore, the work done by the man pushing the car over the given distance is 1000J.

Learn more about work done: brainly.com/question/26115962

7 0
2 years ago
Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of th
Illusion [34]

Complete Question

Apollo 14 astronaut Alan B. Shepard Jr. used an improvised six-iron to strike two golf balls while on the Fra Mauro region of the moon’s surface, making what some consider the longest golf drive in history. Assume one of the golf balls was struck with initial velocity v0 = 32.75 m/s at an angle θ = 32° above the horizontal. The gravitational acceleration on the moon’s surface is approximately 1/6 that on the earth’s surface. Use a Cartesian coordinate system with the origin at the ball's initial position.

Randomized Variables

vo 32.75 m/s

theta 32 degrees

What horizontal distance, R in meters, did this golf ball travel before returning to the lunar surface?

Answer:

The  horizontal distance is  R =  590.2 \ m  

Explanation:

From the question we are told that

 The initial  velocity is  v_o =  32.75 \  m/s

  The angle is  \theta =  26^o

  The gravitational acceleration of the moon is g_m  =  \frac{1}{6}  *  9.8   = 1.633 m/s^2

 Generally the distance traveled is mathematically represented as

    R =  \frac{v_o^2 sin 2(\theta)}{g_m}

=> R =  \frac{32.75^2 sin 2(32)}{1.633}

=> R =  590.2 \ m  

             

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3 years ago
Consider a system to be one train car moving toward another train car at rest. When the train cars collide, the two cars stick t
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Answer:

Momentum after = 2400 kgm/s

Explanation:

Momentum after = momentum before

Momentum after = m1v1 +m2v2

Momentum after = (600)(4) + (400)(0)

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3 years ago
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A Car with drives into a solid object with 70 km/h, from which height would it fall if it was in free fall?​
Ilya [14]

Answer:

18.9 m.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 0 m/s

Final velocity (v) = 70 km/h

Height (h) =?

Next, we shall convert 70 km/h to m/s. This can be obtained as follow:

3.6 km/h = 1 m/s

Therefore,

70 km/h = 70 km/h × 1 m/s / 3.6 km/h

70 km/h = 19.44 m/s

Finally, we shall determine the height. This can be obtained as follow:

Initial velocity (u) = 0 m/s

Final velocity (v) = 19.44 m/s

Acceleration due to gravity (g) = 10 m/s²

Height (h) =?

v² = u² + 2gh

19.44² = 0² + (2 × 10 × h)

377.9136 = 0 + 20h

377.9136 = 20h

Divide both side by 20

h = 377.9136 / 20

h = 18.9 m

Thus, the car will fall from a height of 18.9 m

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What is the energy Q released when 131 53Idecays and 131 54Xe is formed? The atomic mass of 131 53I is 130.906118 u and the atom
DanielleElmas [232]

Answer:0.967meV

Explanation:

-Find the difference in u, so 130.906118-130.90508= 0.001038u

- convert to meV

1 u = 931.494meV

multiply 0.001038 by 931.494

=0.001038 X 931.494

0.967 meV

5 0
4 years ago
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