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Setler [38]
2 years ago
12

A toy car has a momentum of 3 kilogram meters per second south. The car has a 1-kilogram mass. Which is the velocity of the car?

Physics
1 answer:
antiseptic1488 [7]2 years ago
6 0
Using p = mv 3 = 1× v v = 3m/s
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What is the weight of an object when the object has a mass of 22kg
Savatey [412]
Assuming the object is on earth the objects weight would be equal to its mass multiplied by the gravitational field constant

mass=22kg
g=9.80665N/kg

weight=(22 kg) (9.80665 N/kg)=215.7463N

generally g is rounded to be 10 N/kg so for any question where it asks the weight given the mass just multiply by 10 and that should suffice. In this case the answer would be 220 N
5 0
2 years ago
Mr. MacDougall got his vehicle stuck in the snow. Being the nice student that you are, you stop to help Mr. MacDougall out of th
zhannawk [14.2K]

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Explanation:

work done=force*displacement

=350N*15m

=5250 joule

4 0
2 years ago
Waves interact with ___ and other ___.
lutik1710 [3]
Sand and other sources?
5 0
2 years ago
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A circular radar antenna on a coast guard ship has a diameter of 2.10 m and radiates at a frequency of 16.0 ghz. two small boats
solmaris [256]
By definition,
q = 1.22y/D

Where,
q = min. angle
y = wavelength
D = Aperture diameter = diameter of the antenna

At distance "x" from the antenna,
L =xq = 1.22xy/D
Where, L = Min. distance

But, y =c/f = (3*10^8)/(16*10^9) = 0.01875 m

Substituting;
L = 1.22*5*10^3*0.01875/2.1 = 54.46 m
5 0
3 years ago
If 745-nm and 660-nm light passes through two slits 0.54 mm apart, how far apart are the second-order fringes for these two wave
kotegsom [21]

Answer:

0.82 mm

Explanation:

The formula for calculation an n^{th} bright fringe from the central maxima is given as:

y_n=\frac{n \lambda D}{d}

so for the distance of the second-order fringe when wavelength \lambda_1 = 745-nm can be calculated as:

y_2 = \frac{n \lambda_1 D}{d}

where;

n = 2

\lambda_1 = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y_2 = 0.00276 m

y_2 = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength \lambda_2 = 660-nm is as follows:

y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}

y^'}_2 = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

\delta y = y_2-y^{'}_2

\delta y = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

\delta y = 10 ⁻³ (2.76 - 1.94)

\delta y = 10 ⁻³ (0.82)

\delta y = 0.82 × 10 ⁻³ m

\delta y =  0.82 × 10 ⁻³ m (\frac{1.0mm}{10^{-3}m} )

\delta y = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

6 0
3 years ago
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