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just olya [345]
3 years ago
12

An upward force of 27.5 n is applied to a string to lift a ball with a mass of 2.5 kg.

Physics
1 answer:
Rudiy273 years ago
4 0
We first find that weight W of the ball
where W= ma
W= (2.5)(9.81)
= 24.525
therefore the net force acting upon the ball is
= upward force - weight
= 27.5 - 24.525
= 2.975 N
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A shield of material that reflects neutrons is placed around a radioactive sample. Which change in the nuclear reaction is most
Crazy boy [7]

Answer:

A chain reaction will be sustained in a sub-critical mass.

Explanation:

Hope this helps!

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3 0
1 year ago
The total mass of the wheelbarrow and the road is 80 kg calculate the weight of the wheelbarrow and the road
Alja [10]

Answer:

The weight of the wheelbarrow and the road is 784 N and the force required to lift the wheelbarrow is 784 N.

Explanation:

Given that,

The total mass of the wheelbarrow and the road is 80 kg.

The weight of an object is given by :

W = mg

where

g is acceleration due to gravity

So,

W = 80 × 9.8

= 784 N

So, the force required to lift the wheelbarrow is equal to its weight i.e. 784 N.

7 0
3 years ago
A trough is 10 feet long and its ends have the shape of isosceles triangles that are 3 feet across at the top and 1 foot high. I
tensa zangetsu [6.8K]

Answer:

1.6 ft/min

Explanation:

Since trough is 10 ft long and water is filled at the rate of 12ft3/min. We can calculate the rate of water filled with respect to area:

= 12 / 10 = 1.2ft2/min

As the water level rises, so does the water surface, or the bottom side of the isosceles triangles. In fact we can calculate the bottom side when the trough is half foot deep:

= 3 / 2 = 1.5 ft

The rate of change in water level would be the same as calculating the height of the isosceles triangles knowing its base

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6 0
3 years ago
The CERN particle accelerator is circular with a circumference of 7.0 km.
Contact [7]

Answer:

a_c=2.0196\times 10^{13}\ m/s^2

F=3.37273\times 10^{-14}\ N

Explanation:

m = Mass of proton = 1.67\times 10^{-27}\ kg

v = Speed of proton = 0.5c = 0.5\times 3\times 10^8=1.5\times 10^8\ m/s

Circumference of the colider is 7 km

P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m

a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2

Centripetal acceleration is 2.0196\times 10^{13}\ m/s^2

F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N

Force on protons is 3.37273\times 10^{-14}\ N

8 0
3 years ago
A mass of 2 kg traveling at 3 m/s undergoes a one-dimensional elastic collision with a group of four 1kg masses that are at rest
alukav5142 [94]
2kg + 3 kg = 5 kg as result of the conclusion
6 0
3 years ago
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