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Mamont248 [21]
4 years ago
10

A ping-pong ball is shot straight north from a popgun at 4.0 m/s. The wind is blowing to the west at 3.0 m/s. What is the actual

velocity of the ping-pong ball? 7.0 m/s 1.0 m/s
Physics
1 answer:
Vlad1618 [11]4 years ago
5 0

Actual velocity of the ping-pong ball= 5 m/s

Explanation:

velocity of ping pong ball because of the shot gun= 4 m/s North

velocity added to the ping-pong ball due to the wind=3 m/s

These velocities are perpendicular to each other. so we use Pythagoras theorem to find the resultant velocity of the ping- pong ball

so the actual velocity of the ping-pong ball =V= √4²+3²

V= √25

V= 5 m/s

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Answer:

(a) F = 239.575 N (b) t = 0.00649s or 6.49 ms

Explanation:

(a) By law of energy conservation, the bullet kinetic energy will be transferred to work done on stopping it from moving.

Formula for Kinetic Energy E_k = \frac{mv^2}{2} where m is bullet mass, v is the velocity

Formula for work W = FS where F is the average force and S is the distance travelled.

E_k = W

\frac{mv^2}{2} = FS

F = \frac{mv^2}{2S}

Substitute m = 4.2 g = 0.0042 kg, v = 370 m/s and S = 1.2 (m)

F = \frac{0.0042*370^2}{2*1.2} = 239.575 N

(b) If the force is constant, since the mass is constant and F = ma according to Newton's 2nd law, the acceleration on bullet is also constant

a = \frac{F}{m} = \frac{239.575}{0.0042} = 57041.67 (m/s^2)

We also have v(t) = v_0 + at

At the time the bullet is coming to rest, v(t) = 0, a = -57041.67 m/s^2

Therefore, 0 = 370 - 57041.67t

t = \frac{370}{57041.67} = 0.00649 s = 6.49 ms

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What is meant by action and reaction
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Answer: action and reaction forces and are the subject of Newton's third law of motion. Formally stated, Newton's third law is: For every action, there is an equal and opposite reaction

Explanation:

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As the ice melts into liquid water, what happens to the kinetic energy of the particles in the sample?
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A long solenoid with 8.22 turns/cm and a radius of 7.00 cm carries a current of 19.4 mA. A current of 3.59 A exists in a straigh
daser333 [38]

Answer:

a. 3.039cm

b.magnetic field is B=2.958\times10^{-5}T

Explanation:

Direction of the solenoid magnetic field is along the axis of the solenoid. and magnetic field due to the wire perpendicular to that due to the solenoid.. Magnetic field at r is given by:

\overrightarrow B = \overrightarrow B_s+ \overrightarrow B_w,\ \ \ \ \  \overrightarrow B_s\perp \overrightarrow B_w

Angle of net magnetic field from axial direction is given by:

tan\  \theta=\frac{B_w}{B_s},

Field due to solenoid:

B_s=\mu_onI_s,  \ \ \ \ n=(8.22 t/cm)(100cm/m)=822turn/m

Field due to wire:

B_w=\frac{\mu_oI_w}{2\pi r}

Therefore, r:

tan\  \theta=\frac{B_w}{B_s}\\\\=\frac{\mu_oI_w}{2\pi r(\mu_o nI_s)}\\\\r=\frac{I_w}{2\pi  nI_stan \ \theta}\\\\r=\frac{3.59A}{2\pi\times822\times19.4\times10^{-3}A \ tan 49.7\textdegree}\\\\r=3.039cm

Hence, the radial distance is 3.039cm

b.The magnetic field strength is given by:

B=\sqrt{B_w^2+B_s^2}\\\\tan 49.7\textdegree=\frac{B_w}{B_s}\\\\1.179=\frac{B_w}{B_s}\\\\B_w=1.179B_s\\\\B=\sqrt{(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{3}A)+1.179(4\pi\times10^{-7}T.m/A\times 822\times19.4\times10^{-3}A)}\\\\B=2.958\times10^{-5}T

Hence, the magnetic field is B=2.958\times10^{-5}T

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