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2 years ago

A car with a momentum of 3,200 kg-m/s moves foward at a rate of 2 m/s. What is the mass of the vehicle?

Physics

1 answer:

natulia [17]2 years ago

5 0

Answer:

1600 kg

Explanation:

use the formula p=mv. p=3200, v=2. Plug in and rearrange.

3200=(m)(2)

m= 3200/2

m=1600

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taurus [48]

The answer is south

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2 years ago

A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2

Pani-rosa [81]

Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel $I=25kgm^2$

Initial angular velocity of the wheel $\omega _0=10rad/sec$

Angular acceleration $\alpha =15rad/sec^2$

(a) We know that $\omega =\omega _0+\alpha t$

We have given t = 2 sec

So $\omega =10+15\times 2=40rad/sec$

Now $\Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad$

(b) After 3 sec $\omega =10+15\times 3=55rad/sec$

We know that kinetic energy is given by $Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J$

7 0

2 years ago

A current of 4.00 mA flows through a copper wire. The wire has an initial diameter of 4.00 mm which gradually tapers to a diamet

lesya692 [45]

The change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

The given parameters;

Current flowing in the wire, I = 4.00 mA
Initial diameter of the wire, d₁ = 4 mm = 0.004 m
Final diameter of the wire, d₂ = 1 mm = 0.001 m
Length of wire, L = 2.00 m
Density of electron in the copper, n = 8.5 x 10²⁸ /m³

The initial area of the copper wire;

$A_1 = \frac{\pi d^2}{4} = \frac{\pi \times (0.004)^2}{4} =1.257\times 10^{-5} \ m^2$

The final area of the copper wire;

$A_2 = \frac{\pi d^2}{4} = \frac{\pi (0.001)^2}{4} = 7.86\times 10^{-7} \ m^2$

The initial drift velocity of the electrons is calculated as;

$v_d_1 = \frac{I}{nqA_1} \\\\v_d_1 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 1.257\times 10^{-5}} \\\\v_d_1 = 2.34 \times 10^{-8} \ m/s$

The final drift velocity of the electrons is calculated as;

$v_d_2 = \frac{I}{nqA_2} \\\\v_d_2 = \frac{4\times 10^{-3} }{8.5\times 10^{28} \times 1.6\times 10^{-19} \times 7.86\times 10^{-7}} \\\\v_d_2 = 3.74\times 10^{-7} \ m/s$

The change in the mean drift velocity is calculated as;

$\Delta v = v_d_2 -v_d_1\\\\\Delta v = 3.74\times 10^{-7} \ m/s \ -\ 2.34 \times 10^{-8} \ m/s = 3.506\times 10^{-7} \ m/s$

The time of motion of electrons for the initial wire diameter is calculated as;

$t_1 = \frac{L}{v_d_1} \\\\t_1 = \frac{2}{2.34\times 10^{-8}} \\\\t_1 = 8.547\times 10^{7} \ s$

The time of motion of electrons for the final wire diameter is calculated as;

$t_2 = \frac{L}{v_d_1} \\\\t_2= \frac{2}{3.74 \times 10^{-7}} \\\\t_2 = 5.348 \times 10^{6} \ s$

The average acceleration of the electrons is calculated as;

$a = \frac{\Delta v}{\Delta t} \\\\a = \frac{3.506 \times 10^{-7} }{(8.547\times 10^7)- (5.348\times 10^6)} \\\\a = 4.38\times 10^{-15} \ m/s^2$

Thus, the change in mean drift velocity for electrons as they pass from one end of the wire to the other is 3.506 x 10⁻⁷ m/s and average acceleration of the electrons is 4.38 x 10⁻¹⁵ m/s².

Learn more here: brainly.com/question/22406248

7 0

2 years ago

A car moving with an initial speed v collides with a second stationary car that is one-half as massive. After the collision the

Mashutka [201]

Answer:

4v/3

Explanation:

Assume elastic collision by the law of momentum conservation:

$m_1v = m_1v_1 + m_2v_2$

where v is the original speed of car 1, v1 is the final speed of car 1 and v2 is final speed of car 2. m1 and m2 are masses of car 1 and car 2, respectively

Substitute $m_2 = m_1/2 \& v_1 = v/3$

$m_1v = \frac{m_1v}{3} + \frac{m_1v_2}{2}$

Divide both side by $m_1$ , then multiply by 6 we have

$6v = 2v + 3v_2$

$3v_2 = 4v$

$v_2 = \frac{4v}{3}$

So the final speed of the second car is 4/3 of the first car original speed

5 0

2 years ago

Describe what would happen if you rubbed a mineral with a Mohs hardness value of 7 against a mineral with a value of 5?

chubhunter [2.5K]

The mineral with Mohs hardness would be scratched because the mineral with Mohs 7 hardness is stronger than the Mohs 5 mineral. Eventually, that mineral would turn into dust if you kept rubbing it.

8 0

2 years ago