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elena-14-01-66 [18.8K]
3 years ago
9

A 5.00 kilogram block slides along a horizontal,frictionless surface at 10.0 meters per second. for 4.00 seconds. The magnitude

of the blocks momentum is ?
A)200 Kg m/s
B)50.0 Kg m/s
C)20.0 Kg m/s
D)12.5 Kg m/s
Physics
1 answer:
MArishka [77]3 years ago
3 0
It's B. They are trying to trick you by adding the time so you would divide the speed by time, which then you multiply to 5 and get D's answer. Ignore the time. The time does not matter. just multiply 5 and 10 to get 50.

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Aircraft sometimes acquire small static charges. Suppose a supersonic jet has a 0.650 µC charge and flies due west at a speed of
Ganezh [65]

Answer : F=2808\times 10^{-11}\ N

Explanation :

It is given that,

Charge, q=0.650\ \mu C=0.65\times 10^{-6}\ C

Velocity of Aircraft, v=540\ m/s  (in west)

Magnetic field, B=8 \times 10^{-5}\ T ( in north )

Using the relation as :

F=q(v\times B)

Magnetic force is ,

F=0.65\times 10^{-6}\ C\times 540\ m/s\times 8 \times 10^{-5}\ T

F=2808\times 10^{-11}\ N

Using Right hand thumb rule, the direction of force is in the plane perpendicular to the velocity and the magnetic field i.e. -\hat{k}.

4 0
4 years ago
a train travels 81 kilometers in 2 hours and then 88 kilometers in 2 hours what is its average speed​
Blababa [14]

Average speed = total distance / total time

= (81 + 88) / (2 + 2)

= 169/4

= 42.25 kilometers/hour

Let me know if you have any questions.

4 0
3 years ago
A traveling electromagnetic wave in a vacuum has an electric field amplitude of 50.9 V/m. Calculate the intensity ???? of this w
Sliva [168]

Answer:

3.44 W/m²

1.134 J

Explanation:

E₀ = Intensity of electric field = 50.9 V/m

I = Intensity of electromagnetic wave

Intensity of electromagnetic wave is given as

I = (0.5) ε₀ E₀² c

I = (0.5) (8.85 x 10⁻¹²) (50.9)² (3 x 10⁸)

I = 3.44 W/m²

A = Area = 0.0277 m²

t = time interval = 11.9 s

Amount of energy is given as

U = I A t

U = (3.44) (0.0277) (11.9)

U = 1.134 J

5 0
3 years ago
A person walks into a room that has, on opposite walls, two plane mirrors producing multiple images. Find the distances from the
Lapatulllka [165]

Answer:13.2,41.4,82.8 ft

Explanation:

Given Person is 6.60 ft from the left hand side mirror

Since the focal length of plane mirror is infinity therefore image and object are equidistant from mirror

Distance of first image on  left mirror is

d_1=6.6+6.6=13.2 ft

i.e. image is 13.2 ft away from object

Second image

Now the right mirror forms the image of object at distance of 14.1 ft right from right mirror so its image is formed in left mirror at a distance of 34.8 from left mirror

so image distance from person is

d_2=34.8+6.6=41.4 ft

Third image now form image of second image is formed on right mirror at a distance of 55.5 right from right mirror

and its mirror image is formed on left mirror at a distance of  76.2 left

from left mirror

d_3=76.2+6.6=82.8 ft

     

8 0
3 years ago
A voltmeter has an internal resistance of 10 000 Ω and is used to measure the voltage across a 47.0-Ω resistor that, without the
otez555 [7]

Answer:

b. 5.6 mA

Explanation:

As the voltmeter is connected to the resistor in parallel. The new resistance of the systems is

\frac{1}{R} = \frac{1}{r_1} + \frac{1}{r_2} = \frac{1}{10000} + \frac{1}{47} = 0.0214

R = 1/0.0214 = 46.78 Ω

The voltage is

U = r_2 I_1 = 47 * 1.2 = 56.4 V

So the new current now of the system is

I_2 = U/R = 56.4/46.78 = 1.2056 A

So about 1.2056 - 1.2 = 0.0056 A or 5.6mA is drawn away.

7 0
4 years ago
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