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Anuta_ua [19.1K]
3 years ago
12

Which of the following is the correct Ohm's law formula?

Physics
1 answer:
Mariulka [41]3 years ago
3 0

Answer:

The answer is B .pls don't forget to hit me brainliest tnks

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3. What is the acceleration of a 10 kg mass pushed by a 5 N force?
insens350 [35]

Answer:

F=ma

Plug it in:

5=10a

5/10=(10a)/10

.5m/s²=a

Explanation:

Brainliest?

5 0
3 years ago
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A loader sack of total mass
vampirchik [111]

Question: A loader sack of total mass

is l000 grams falls down from

the floor of a lorry 200 cm high

Calculate the workdone by the

gravity of the load.​

Answer:

19.6 Joules

Explanation:

Applying

W = mgh........................ Equation 1

Where W = Workdone by gravity on the load, m = mass of the loader sack, h = height, g = acceleration due to gravity

From the question,

Given: m = 1000 grams = (1000/1000) kilogram = 1 kg, h = 200 cm = 2 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

W = (1×2×9.8)

W = 19.6 Joules

Hence the work done by gravity on the load is 19.6 Joules

8 0
3 years ago
What is the value of the normal force if the coefficient of kinetic friction is 0.22 and the kinetic frictional force is 40 newt
Salsk061 [2.6K]
1.8x10 sqaured newtons for all plato users

5 0
3 years ago
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Two spheres are cut from a certain uniform rock. One has radius 4.50 cm. The mass of the other is five times greater. Find its r
Sonja [21]
Vol of sphere = 4/3 pi r^2.density of sphere = mass/volume.mass = densityxvolumesphere 1. mass = density x 4/3 pi 4.5^2sphere 2 5mass = density x 4/3 pi r^25=4/3 pi r^2 divided by 4/3 pi 4.5^25=r^2 divided by  4.5^25x4.5^2=r^2root(5x4.5^2)=r4.5 root 5 = r
4 0
2 years ago
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Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of
vredina [299]

Answer:

The angular speed after 6s  is \omega = 1466.67s^{-1}.

Explanation:

The equation

I\alpha  = Fd

relates the moment of inertia I of a rigid body, and its angular acceleration \alpha, with the force applied F at a distance d from the axis of rotation.

In our case, the force applied is F = 22N, at a distance d = 6cm =0.06m, to a ring with the moment of inertia of I =mr^2; therefore, the angular acceleration is

$\alpha =\frac{Fd}{I} $

$\alpha =\frac{22N*0.06m}{(1.5kg)*(0.06)^2} $

\alpha  = 244.44\: s^{-2}

Therefore, the angular speed \omega which is

\omega  = \alpha t

after 6 seconds is

\omega = 244.44$\: s^{-2}* 6s

\boxed{\omega = 1466.67s^{-1}}

7 0
3 years ago
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