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3 years ago

If a car is traveling 60 mph, its tires may have an rpm of 840. How many revolutions do the car’s tires make in one second?

Physics

1 answer:

Varvara68 [4.7K]3 years ago

5 0

Forget about the car's speed. You don't need it.

The tires spin 840 rpm. That's 840 Revolutions per Minute .

There are 60 seconds in 1 minute. So something that happens 840 times in 1 minute happens (840 / 60) times every second.

(840 rev per minute) / (60) = 14 revs per second .

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It takes a minimum distance of 57.46 m to stop a car moving at 13.0 m/s by applying the brakes (without locking the wheels). Ass

vivado [14]

Answer:

The minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

Explanation:

We know by equation of motion that,

$v^{2}=u^{2}+2\cdot a \cdot s$

Where, v= final velocity m/sec

u=initial velocity m/sec

a=Acceleration m/ $Sec^{2}$

s= Distance traveled before stop m

Case 1

u= 13 m/sec, v=0, s= 57.46 m, a=?

$0^{2} = 13^{2} + 2 \cdot a \cdot57.46$

a = -1.47 m/ $Sec^{2}$ (a is negative since final velocity is less then initial velocity)

Case 2

u=29 m/sec, v=0, s= ?, a=-1.47 m/ $Sec^{2}$ (since same friction force is applied)

$v^{2} = 29^{2} - 2 \cdot 1.47 \cdot S$

s = 285.94 m

Hence the minimum stopping distance when the car is moving at

29.0 m/sec = 285.94 m

4 0

3 years ago

What's the difference between electromagnetic waves and electromagnetic spectrum?

White raven [17]

The difference between the two is, well for one

Spectrum: The entire range that the "waves" could be such, as visible light, x-ray's and so on.

Waves: These are different because they aren't telling you or showing the entire spectrum just which they length that they are.

It may confuse you but it makes sense to me (Sorry)

4 0

3 years ago

the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m

olga_2 [115]

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

$x = \frac {-nmg}{k}$

Substituting into the formula, we have;

$x = \frac {-49*0.002*9.8}{24}$

$x = \frac {-0.9604}{24}$

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

3 0

3 years ago

The equation T^2=A^3 shows the relationship between a planet’s orbital period, T, and the planet’s mean distance from the sun, A

kobusy [5.1K]

Answer:

D. 2^(3/2)

Explanation:

Given that

T² = A³

Let the mean distance between the sun and planet Y be x

Therefore,

T(Y)² = x³

T(Y) = x^(3/2)

Let the mean distance between the sun and planet X be x/2

Therefore,

T(Y)² = (x/2)³

T(Y) = (x/2)^(3/2)

The factor of increase from planet X to planet Y is:

T(Y) / T(X) = x^(3/2) / (x/2)^(3/2)

T(Y) / T(X) = (2)^(3/2)

3 0

3 years ago

A point source of light is located at the bottom of a steel tank, and an opaque circular card of radius is placed horizontally o

pochemuha

Start by using the addition as a sign and use multiplying

3 0

3 years ago