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Valentin [98]
3 years ago
10

I’m so confused. please help. i don’t know what i’m suppose to do

Physics
1 answer:
aniked [119]3 years ago
4 0

Answer:

Maybe put them in order ????

Explanation:

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The slope of the line on a speed-time graph tells the speed.
iren2701 [21]

From the choices provided, the better answer is ' T ' .

8 0
3 years ago
How do you do this question? Please include free body diagrams and clear explanation, so I can understand.
vagabundo [1.1K]

Explanation:

Draw a free body diagram for each disc.

Disc A has three forces acting on it: 86.5 N up, T₁ down, and Wa down.

∑F = ma

86.5 N − T₁ − Wa = 0

Wa = 86.5 N − T₁

ma × 9.8 m/s² = 86.5 N − 55.6 N

ma = 3.2 kg

Disc B has three forces acting on it: T₁ up, T₂ down, and Wb down.

∑F = ma

T₁ − T₂ − Wb = 0

Wb = T₁ − T₂

mb × 9.8 m/s² = 55.6 N − 36.5 N

mb = 1.9 kg

Disc C has three forces acting on it: T₂ up, T₃ down, and Wc down.

∑F = ma

T₂ − T₃ − Wc = 0

Wc = T₂ − T₃

mc × 9.8 m/s² = 36.5 N − 9.6 N

mc = 2.7 kg

Disc D has two forces acting on it: T₃ up and Wd down.

∑F = ma

T₃ − Wd = 0

Wd = T₃

md × 9.8 m/s² = 9.6 N

md = 0.98 kg

3 0
3 years ago
How much watt is called 1 H. P​
agasfer [191]

Answer:

746 watts

Explanation:

7 0
3 years ago
Hooke's law describes a certain light spring of unstretched length 34.8 cm. When one end is attached to the top of a doorframe a
DaniilM [7]

Answer:

a) k = 1343.6\,\frac{N}{m}, b) l = 0.501\,m\,(50.1\,cm)

Explanation:

a) The Hooke's law states that spring force is directly proportional to change in length. That is to say:

F \propto \Delta l

In this case, the force is equal to the weight of the object:

F = m\cdot g

F = (8.22\,kg)\cdot (9.807\,\frac{m}{s^{2}} )

F = 80.614\,N

The spring constant is:

k = \frac{F}{\Delta l}

k = \frac{80.614\,N}{0.408\,m-0.348\,m}

k = 1343.6\,\frac{N}{m}

b) The length of the spring is:

F = k\cdot (l-l_{o})

l = l_{o} + \frac{F}{k}

l=0.348\,m+\frac{205\,N}{1343.6\,\frac{N}{m} }

l = 0.501\,m\,(50.1\,cm)

6 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
4 years ago
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