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2 years ago

Two cars move down a hill at a constant velocity. Car c is much larger than car d. The car that has a greater momentum is:

Physics

2 answers:

babunello [35]2 years ago

7 0

Answer:

Explanation:

When an object has a greater mass than the drag force pushing it back then it is moving forward therefore making car C faster than car D. If car D was much smaller it wouldn't have enough mass to push past the drag force.

BlackZzzverrR [31]2 years ago

4 0

The answer would be c

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A crate with a mass of m = 450 kg rests on the horizontal deck of a ship. The coefficient of static friction between the crate a

Zielflug [23.3K]

Answer: $F_{v} =\mu_{k} mg$

Magnitude of the force is 2601.9 N

Explanation:

m = 450 kg

coefficient of static friction μs = 0.73

coefficient of kinetic friction is μk = 0.59

The force required to start crate moving is $F_{s} =\mu_{s} mg$ .

but once crate starts moving the force of friction is reduced $F_{v} =\mu_{k} mg$ .

Hence to keep crate moving at constant velocity we have to reduce the force pushing crate ie $F_{v} =\mu_{k} mg$ .

Then the above pushing force will equal the frictional force due to kinetic friction and constant velocity is possible as forces are balanced.

Magnitude of the force

$F_{v} =\mu_{k} mg\\F_{v} =0.59 \times 450 \times 9.8\\F_{v} =2601.9 N$

4 0

3 years ago

A positive test charge of 8.5 × 10 negative 7 Columbus experiences a force of 4.1 × 10 negative 1 N calculate the electric field

Sophie [7]

Explanation:

direction of electric field is same as that of force experienced by the test charge

8 0

2 years ago

A radar for tracking aircraft broadcasts a 12 GHz microwave beam from a 2.0-m-diameter circular radar antenna. From a wave persp

NISA [10]

A) 750 m

First of all, let's find the wavelength of the microwave. We have

$f=12GHz=12\cdot 10^9 Hz$ is the frequency

$c=3.0\cdot 10^8 m/s$ is the speed of light

So the wavelength of the beam is

$\lambda=\frac{c}{f}=\frac{3\cdot 10^8 m/s}{12\cdot 10^9 Hz}=0.025 m$

Now we can use the formula of the single-slit diffraction to find the radius of aperture of the beam:

$y=\frac{m\lambda D}{a}$

where

m = 1 since we are interested only in the central fringe

D = 30 km = 30,000 m

a = 2.0 m is the aperture of the antenna (which corresponds to the width of the slit)

Substituting, we find

$y=\frac{(1)(0.025 m)(30000 m)}{2.0 m}=375 m$

and so, the diameter is

$d=2y = 750 m$

B) 0.23 W/m^2

First we calculate the area of the surface of the microwave at a distance of 30 km. Since the diameter of the circle is 750 m, the radius is

$r=\frac{750 m}{2}=375 m$

So the area is

$A=\pi r^2 = \pi (375 m)^2=4.42\cdot 10^5 m^2$

And since the power is

$P=100 kW = 1\cdot 10^5 W$

The average intensity is

$I=\frac{P}{A}=\frac{1\cdot 10^5 W}{4.42\cdot 10^5 m^2}=0.23 W/m^2$

4 0

2 years ago

1-Find the gravitational force

butalik [34]

$g = \frac{GMm}{r^2}$

where $G$ is the gravitational constant and is about $6.67 * 10^{-11}$

$g = \frac{6.67 * 10^{-11} * 25 * 0.55}{35^2}$

$g = \frac{6.67 * 10^{-11} * 25 * 0.55}{1225} = 0.000000000000749 N$ :D

5 0

2 years ago

Which interaction of sound waves causes beats? (Apex)

liubo4ka [24]

The answer to this question would be A. Interference. This is because when two sounds with slightly different frequencies creates a periodic variation in volume.

5 0

2 years ago

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