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frez [133]
3 years ago
14

A 27.0 g marble sliding to the right at 56.8 cm/s overtakes and collides elastically with a 13.5 g marble moving in the same dir

ection at 14.2 cm/s. After the collision, the 13.5 g marble moves to the right at 71 cm/s. Find the velocity of the 27.0 g marble after the collision.
Physics
1 answer:
Nataly [62]3 years ago
7 0

Answer:v_1=28.4 cm/s

Explanation:

Given

mass of marble m_1=27 gm

velocity of marble u_1=56.8 cm/s \approx 0.568 m/s

mass of second marble m_2=13.5 gm

Velocity of second marble u_2=14.2 cm/s \approx 0.142 m/s

After collision 13.5 gm marble moves to the right  at i.e. v_2=71 cm/s

Conserving momentum

m_1u_1+m_2u_2=m_1v_1+m_2v_2

27\times 56.8+13.5\times 14.2=27\times v_1+13.5\times 71

1533.6+191.7=27\cdot v_1+958.5

27\cdot v_1=766.8

v_1=\frac{766.8}{27}=28.4 cm/s

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svlad2 [7]

Answer:

Woke done, W = 4156.92 Joules

Explanation:

The work done by the force can be calculated as :

W=F\times s

W=Fs\ cos\theta

\theta is the angle between force and the displacement

It is assumed to find the work done for the given parameters i.e.

Force, F = 30 N

Distance travelled, s = 160 m

Angle between force and displacement, \theta=30

Work done is given by :

W=Fs\ cos\theta

W=30\times 160\ cos(30)

W = 4156.92 Joules

So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.

5 0
2 years ago
An object starts from rest, and accelerates at 2m/s2 for 10s. How far has it gone in that time
blagie [28]

Answer:

100m

Explanation:

s = ut +  \frac{1}{2} a {t}^{2}

u=0;t=10sec;a=2m/s²

s = 0(10) +  \frac{1}{2}(2 \times  {10}^{2} )

s=10²;100m

7 0
3 years ago
PLEASEE HELP!!! LOTS OF POINTS AND BRAINLIEST!!!!!!!!
Dennis_Churaev [7]
A clock and a battery
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2 years ago
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A mass of 0.54 kg attached to a vertical spring stretches the spring 36 cm from its original equilibrium position. The accelerat
UNO [17]
<h2>Spring constant is 14.72 N/m</h2>

Explanation:

We have for a spring

            Force =  Spring constant x Elongation

            F = kx

Here force is weight of mass

           F = W = mg = 0.54 x 9.81 = 5.3 N

Elongation, x  = 36 cm = 0.36 m

Substituting

           F = kx

           5.3 = k x 0.36

             k = 14.72 N/m

Spring constant is 14.72 N/m

6 0
2 years ago
A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

\dfrac{n}{t}=16250000

The number of electron passe per second will be 162500000.

4 0
2 years ago
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