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OlgaM077 [116]
3 years ago
13

A 15kg projectile is moving with a velocity of 25m/s calculate its momentum

Physics
1 answer:
aliina [53]3 years ago
8 0

The momentum of an object is given by:

p = mv

m is the object's mass and v is its velocity.

Given values:

m = 15kg

v = 25m/s

Plug in the values and solve for p:

p = 15*25

p = 375kg×m/s

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Two balls of clay, with masses M1 = 0.49 kg and M2 = 0.47 kg, are thrown at each other and stick when they collide. Mass 1 has a
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Answer:

a) p_i=1.568\hat{i}+0.752 \hat{j}

b) v_{fx}=1.668\ m.s^{-1}

c) v_{fy}=0.7999\ m.s^{-1}

Explanation:

Given masses:

m_1=0.49\ kg

m_2=0.47\ kg

Velocity of mass 1, v_1=3.2 \hat{i}\ m.s^{-1}

Velocity of mass 2, v_2=1.6 \hat{j}\ m.s^{-1}

a)

Initial momentum:

p_i=m_1.v_1+m_2.v_2

p_i=0.49\times 3.2 \hat{i}+0.47\times 1.6 \hat{j}

p_i=1.568\hat{i}+0.752 \hat{j}

b)

magnitude of initial momentum:

p_i=\sqrt{1.568^2+0.752 ^2}

p_i=1.739\ kg.m.s^{-1}

From the conservation of momentum:

p_f=p_i

m_f.v_f=1.739

v_f=\frac{1.739}{0.49+0.47}

v_f=1.85\ m.s^{-1} is the magnitude of final velocity.

Direction of final velocity will be in the direction of momentum:

tan\theta=\frac{0.752 }{1.568}

\theta=25.62^{\circ}

\therefore v_{fx}=1.85\ cos25.62^{\circ}

v_{fx}=1.668\ m.s^{-1}

c)

Vertical component of final velocity:

v_{fy}=1.85\ sin 25.62^{\circ}

v_{fy}=0.7999\ m.s^{-1}

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3 years ago
In treatment the so called "nervous " disorder, freuad discovered the most effetive method to be
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Answer:

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Explanation:

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3 years ago
The city of Istanbul turkey sits on what?
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Istanbul is a transcontinental city in Eurasia, straddling the Bosporus strait (which separates Europe and Asia) between the Sea of Marmara and the Black Sea. Its commercial and historical center lies on the European side and about a third of its population lives in suburbs on the Asian side of the Bosporus.

8 0
3 years ago
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An airplane is at rest on a runway. It accelerates at 10 m/s2 for 15 seconds. How fast is it now traveling?​
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Answer:

150

Explanation:

v = at

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6 0
3 years ago
A solid sphere with a radius of 50.0 cm has a total positive charge of 40.0 μC uniformly distributed in its volume. Calculate th
il63 [147K]

Answer:

E=2.88\times 10^5\ N/C                              

Explanation:

It is given that,

The radius of the solid sphere, R = 50 cm = 0.5 m

Charge on the sphere, q=40\ \mu C=40\times 10^{-6}\ C

We need to find the magnitude of the electric field r = 10.0 cm away from the center of the sphere. The electric field at point r away form the center of the sphere is given by :

E=\dfrac{kqr}{R^3}

E=\dfrac{9\times 10^9\times 40\times 10^{-6}\times 0.1}{0.5^3}

E=2.88\times 10^5\ N/C

So, the electric field 10.0 cm away from the center of the sphere is 2.88\times 10^5\ N/C. Hence, this is the required solution.              

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3 years ago
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