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bazaltina [42]
3 years ago
7

14. Saeed is pulling a 6 kg heavy rock with an upward force of 40 N but does not succeed to lift it up. What is the magnitude of

the normal force on the rock?
Physics
1 answer:
NemiM [27]3 years ago
6 0

Answer:

58.8 N

Explanation:

The normal force is calculated as equal to the perpendicular component of the gravitational force.

Thus; N = mg

We are given m = 6 kg

Thus;

N = 6 × 9.8

N = 58.8 N

Thus, magnitude of normal force on the rock = 58.8 N

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A 9.5 V battery supplies a 3.5 mA current to a circuit for 7.0 h. Part A How much charge has been transferred from the negative
pychu [463]

Answer:

88.2 C

Explanation:

The current can be defined as the rate of flow of charge in a conductor.

The relation between charge current and time is given as

I = Q/T

I = current, Q= charge and T = time

that is ampere = coulomb / second

The amount of charge passed is from the negative to the positive terminal

shall be given by:

Q = I * t = 3.5mA * 7h * 3600s/h = 88.2 C

Note: take care of the units.  

7 0
3 years ago
A coin is dropped. What is its displacement after 0.30 s.
Morgarella [4.7K]
Assuming this coin is on earth and that it wasn’t dropped forcefully:
Use the formula d = 1/2at^2. Rewriting using a=g and solving for height h gets us h = 1/2(9.8)t^2.
In this case that would get that the change in height h is 0.5(9.8)(0.3^2) = 0.441 m.
3 0
3 years ago
Explain why offspring may or may not resemble either parent
katen-ka-za [31]
The offspring can have some features for the parents relatives and can look nothing like the parents. They can look exactly alike to more of one parent then the other or have features from both parents as well
Hope this helps :3
6 0
3 years ago
Read 2 more answers
A horizontal force of magnitude 30.2 N pushes a block of mass 3.50 kg across a floor where the coefficient of kinetic friction i
marshall27 [118]

Answer:

A) 89.39 J

B) 30.39J

C) 23.8 J

Explanation:

We are given;

F = 30.2N

m = 3.5 kg

μ_k = 0.646

d = 2.96m

ΔEth (Block) = 35.2J

A) Work done by the applied force on the block-floor system is given as;

W = F•d

Thus, W = 30.2 x 2.96 = 89.39 J

B) Total thermal energy dissipated by the whole system which includes the floor and the block is given as;

ΔEth = μ_k•mgd

Thus, ΔEth = 0.646 x 3.5 x 9.8 x 2.96 = 65.59J

Now, we are given the thermal energy of the block which is ΔEth (Block) = 35.2J.

Thus,

ΔEth = ΔEth (Block) + ΔEth (floor)

Thus,

ΔEth (floor) = ΔEth - ΔEth (Block)

ΔEth (floor) = 65.59J - 35.2J = 30.39J

C) The total work done is considered as the sum of the thermal energy dissipated as heat and the kinetic energy of the block. Thus;

W = K + ΔEth

Therefore;

K = W - ΔEth

K = 89.39 - 65.59 = 23.8J

3 0
3 years ago
A powerful searchlight shines on a man. The man's cross-sectional area is 0.500m2 perpendicular to the light beam, and the inten
babymother [125]

Answer:

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵N

(b) the force the light beam exerts is much too small to be felt by the man.

Explanation:

Given;

cross-sectional area of the man, A = 0.500m²

intensity of light, I = 35.5kW/m²

If all the incident light were absorbed, the pressure of the incident light on the man can be calculated as follows;

P = I/c

where;

P is the pressure of the incident light

I is the intensity of the incident light

c is the speed of light

P = \frac{I}{c} =\frac{35500}{3*10^8} = 1.18*10^{-4} \ N/m^2

F = PA

where;

F is the force of the incident light on the man

P is the pressure of the incident light on the man

A is the cross-sectional area of the man

F = 1.18 x 10⁻⁴ x 0.5 = 5.9 x 10⁻⁵ N

The magnitude of the force the light beam exerts on the man is 5.9 x 10⁻⁵ N

Therefore, the force the light beam exerts is much too small to be felt by the man.

8 0
3 years ago
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