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____ [38]
2 years ago
14

Help hurry help pls it’s timed

Physics
2 answers:
RoseWind [281]2 years ago
5 0

I think it's the yellow one

Jobisdone [24]2 years ago
4 0

Answer:

its the teal/second from the left one

Explanation:

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Timers at a swim meet used four different clocks to time an event which recorded time is the most precise
tatiyna
The answer is b 55.2 that is correct
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3 years ago
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A man on the moon with a mass of 90 kilograms weighs 146 newtons. The radius of the moon is 1.74 x 106 meters, find the mass of
melisa1 [442]

Answer:

7.36 × 10^22 kg

Explanation:

Mass of the man = 90kg

Weight on the moon = 146N

radius of the moon =1.74×10^6

Weight =mg

g= weight/mass

g= 146/90 = 1.62m/s^2

From the law of gravitational force

g = GM/r^2

Where G = 6.67 ×10^-11

M = gr^2/G

M= 1.62 × (1.74×10^6)^2/6.67×10^-11

= 4.904×10^12/6.67×10^-11

=0.735×10^23

M= 7.35×10^22kg. (approximately) with option c

3 0
3 years ago
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The static frictional force between a 95-kilogram object and the floor is 45 Newtons. The kinetic frictional force is only 22 Ne
Lisa [10]

Answer:

F = 69.5 [N]

Explanation:

We must remember that the friction force is defined as the product of the normal force by the coefficient of friction, and it can be calculated by the following expression.

f=N*miu

where:

N = normal force [N]

miu = friction coefficient

f = friction force = 22 [N]

Now we must calculate the force exerted by means of Newton's second law which tells us that the sum of forces on a body is equal to the product of mass by acceleration.

F - f = m*a

where:

F = force exerted [N]

f = friction force [N]

m = mass = 95 [kg]

a = acceleration = 0.5 [m/s²]

Now replacing:

F - 22 = 95*0.5\\F = 47.5 + 22\\F = 69.5 [N]

6 0
3 years ago
A student measures the speed of a rolling ball three times. She adds the measurements and divides by 3. What quantity did the st
Lapatulllka [165]
They calculated the mean value
7 0
3 years ago
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Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat
Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e \frac{dv}{dt} =10 cm^3/s

Consider r be the radius of the balloon.

The volume of of a sphere is

v=\frac{4}{3} \pi r^3

Differentiate with respect to t

\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}

\Rightarrow 10 =4\pi r^2\frac{dr}{dt}

\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}

The surface of area of the balloon is(S) = 4\pi r^2

S=4\pi r^2

Differentiate with respect to t

\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}

Putting the value of \frac{dr}{dt}

\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}

\Rightarrow \frac{dS}{dt} =\frac{20}{ r}

Given that r = 5 cm

[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}  =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0
3 years ago
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