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OlgaM077 [116]
3 years ago
14

A 5.00 kilogram mass is traveling at 100. meters per second. Determine the speed of the mass after an impulse with a magnitude o

f 30.0 newton • seconds is applied.
Physics
2 answers:
faltersainse [42]3 years ago
5 0

m = mass = 5 kg

v_{i} = initial velocity = 100 m/s

v_{f} = final velocity = ?

I = impulse = 30 Ns

Using the impulse-change in momentum equation

I = m(v_{f} - v_{i})

30 = 5 (v_{f} - 100)

v_{f} = 106 m/s

Romashka [77]3 years ago
5 0
The answer to this question is <span>106 m/s</span>
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A solenoid is designed to produce a magnetic field of 3.50×10^−2 T at its center. It has a radius of 1.80 cm and a length of 46.
Anna11 [10]

Answer:

a. 2143 turns/m

b. 111.5 m

Explanation:

a. The minimum number of turns per unit length (N/L) can be found using the following equation:

B = \frac{\mu_{0}NI}{L}

\frac{N}{L} = \frac{B}{\mu_{0}I} = \frac{3.50 \cdot 10^{-2} T}{4\pi \cdot 10^{-7} Tm/A*13.0 A} = 2143 turns/m

Hence, the minimum number of turns per unit length is 2143 turns/m.

b. The total length of wire is the following:

N = 2143 turns/m*L = 2143 turns/m*46.0 \cdot 10^{-2} m = 986 turns

Since each turn has length 2πr of wire, the total length is:

L_{T} = N*2\pi r = 986 turn*2*\pi*1.80 \cdot 10^{-2} m = 111.5 m

Therefore, the total length of wire required is 111.5 m.

I hope it helps you!

4 0
3 years ago
A man starts from rest and accelerates at 4.00 m/s2. If he covers a distance of 525 m, how long does he accelerate?
rosijanka [135]

Answer:

16.2 s

Explanation:

Given:

Δx = 525 m

v₀ = 0 m/s

a = 4.00 m/s²

Find: t

Δx = v₀ t + ½ at²

525 m = (0 m/s) t + ½ (4.00 m/s²) t²

t = 16.2 s

5 0
2 years ago
a train car of mass 444 kg moving at 5 m/s bounces into another car on the same tracks of mass 344 king. of the second car was m
Kamila [148]

Answer:

3.7 m/s

Explanation:

M = 444 kg

U = 5 m/s

m = 344 kg

u = - 5 m/s

Let the velocity of train is V and the car s v after the collision.

As the collision is elastic

By use of conservation of momentum

MU + mu = MV + mv

444 x 5 - 344 x 5 = 444 V + 344 v

500 = 444 V + 344 v

125 = 111 V + 86 v .... (1)

By using the formula of coefficient of restitution ( e = 1 for elastic collision)

e = \frac{V-v}{u-U}

-5 - 5 = V - v

V - v = - 10

v = V + 10

Substitute the value of v in equation (1)

125 = 111 V + 86 (V + 10)

125 = 197 V + 860

197 V = - 735

V = - 3.7 m/s

Thus, the speed of first car after collision is 3.7 m/s. negative sign shows that the direction is reverse as before the collision.

4 0
3 years ago
A +12 μC charge and -8 μC charge are 4 cm apart. Find the magnitude and direction of the E-field at the point midway between t
Natasha_Volkova [10]

Answer:

Explanation:

Given

Charge of first Particle q_1=+12\ \mu C

Charge of second Particle q_2=-8\ \mu C

distance between them d=4\ cm

k=9\times 10^{9}

magnetic field due to first charge at mid-way between two charged particles is

E_1=\frac{kq_1}{r^2}

r=\frac{d}{2}=\frac{4}{2}=2\ cm

E_1=\frac{9\times 10^9\times 12\times 10^{-6}}{(2\times 10^{-2})^2}

E_1=27\times 10^7\ N/C (away from it)

Electric field due to q_2=-8\ \mu C

E_2=\frac{kq_2}{r^2}

E_2=-\frac{9\times 10^9\times 8\times 10^{-6}}{(2\times 10^{-2})^2}

E_2=-18\times 10^7\ N/C(towards it)

E_{net}=E_1+E_2

E_{net}=9\times 10^7\ N/C(away from first charge)        

8 0
3 years ago
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Answer:

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