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Kazeer [188]
3 years ago
7

A ball is thrown straight up with an initial speed of 50 ms how high does it go?

Physics
1 answer:
gavmur [86]3 years ago
6 0
h_{max}=\frac{v_0^2}{2g}=125m (g=10\frac{m}{s^2})
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A car slows down from 21 m/s to rest in a distance of 63m. Assuming the car has a constant acceleration, calculate the time it t
vladimir2022 [97]

Answer:

-3.5 m/s²

Explanation:

  • Initial Velocity = 21m /s
  • Final velocity = 0m/s
  • Distance = 63 m .
  • Acclⁿ = ?

<u>We </u><u>know</u><u> </u><u>that</u><u> </u><u>:</u><u>-</u><u> </u>

\longrightarrow Stopping distance = u²/2(-a)

\longrightarrow 63m = (21m/s)² / -2a

\longrightarrow a = - 21 * 21 / 63 * 2 m/s²

\longrightarrow a = - 3.5 m/s²

<em>*</em><em>*</em><em>Edits</em><em> </em><em>are</em><em> </em><em>welcomed</em><em>*</em><em>*</em>

5 0
3 years ago
If A and B are two objects with masses 6 kg and 34 kg respectively then
Mademuasel [1]

Answer:

the object B has more mass (= 34 kg) than the object A (= 6 kg)................................ ᕦ( ᐛ )ᕡ

4 0
3 years ago
Find the net force for <br> 10 N<br> 10 N<br> 25degree<br> 5N<br> 5N
icang [17]

Answer:

30n

Explanation:

7 0
3 years ago
Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same am
Tcecarenko [31]

Answer:

Incomplete question,

This is the complete question

Suppose you push horizontally with precisely enough force to make the block start to move, and you continue to apply the same amount of force even after it starts moving. Find the acceleration ~a of the block after it begins to move. Express your answer in terms of some or all of the variables µs, µk, and m, as well as the acceleration due to gravity g.

Explanation:

Let the force that make the object to start moving be F,

Frictional force is opposing the motion, the body has to overcome two frictional forces acting in the opposite direction of the motion.

Also, weight and normal reaction are acting in vertical direction, the weight is acting downward while the reaction is acting upward.

Weight of the object is given as

W=mg

Analyzing the vertical motion i.e y-axis.

ΣF = ma

since the body is not moving upward, the a=0

N-W=0

Then, N=W

So, N=mg

So, from friction law

Fr=µN

For static

Fs=µsN

For kinetic or dynamic

Fk= µkN

Using newton law

Along x-axis

Before the body start moving we can get the Force and since the force is the same use to start the block in motion.

Then,

ΣF = ma

Since at static the body is not moving then, a=0

F-Fs=0

F=Fs

Since, Fs=µsN

F=Fs=µsN

Then, the force to keep the body in motion too is F=µsN

Now analyses when the body is in motion

ΣF = ma

F-Fk=ma

ma=F - Fk

Substituting F=µsN and Fk=µkN

ma=µsN - µkN

ma=N(µs - µk)

Since N=mg

Then, ma=mg(µs - µk)

m cancels out, then

a=g(µs - µk)

Then the acceleration of the body is given as "a=g(µs - µk)"

5 0
3 years ago
A car is traveling at 114 m/s and changes its velocity to 77 m/s in 9 sec.
suter [353]

Answer:

<u>Given</u><u> </u><u>-</u>

  • Initial Velocity, u = 114 m/s
  • Final velocity, v = 77 m/s.
  • Time taken, t = 9 sec.

<u>To</u><u> </u><u>find</u><u> </u><u>-</u><u> </u>

  • Acceleration of the car.

<u>Solu</u><u>tion</u><u> </u><u>-</u>

Here, using the equation of motion v = u + at we can find the acceleration easily.

★ Here,

  1. V = Final velocity
  2. U = Initial Velocity
  3. A = Acceleration
  4. T = Time.

<u>Subs</u><u>tituting</u><u> </u><u>the</u><u> </u><u>values</u><u> </u><u>-</u>

→ 77 = 114 + a(9)

→ 9a = 114 - 77

→ 9a = 37

→ a = 37/9

→ a = 4.1 m/s

<u>There</u><u>fore</u><u>,</u><u> </u><u>the</u><u> </u><u>accele</u><u>ration</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>car</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>4</u><u>.</u><u>1</u><u> </u><u>m</u><u>/</u><u>s</u><u>.</u>

5 0
3 years ago
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