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3 years ago

When you stretch a spring 33 cm past its natural length, it exerts a force of 6 N. What is the spring constant of this spring?

2 answers:

6 0

The spring formula:

   Force = - (spring constant) · (distance stretched or squashed)

   6 N = - (spring constant) · (33 cm)

   = - (0.33 m) · (spring constant)

Spring constant = - (6 N) / (0.33 m)

=    - 18.18 N/meter .

The negative signs means that the spring always wants to go in the
opposite direction from where you forced it, and return to its relaxed
length. If you stretch it, it tries to get shorter, and if you compress it,
it tries to get longer.

inessss [21]3 years ago

4 0

Answer:

its 0.18 N/cm

in apeeex!!!

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Which of the circuit offers greater resistance to the flow of current 1A or 2A ?

Dafna1 [17]

1 A offers greater resistance

7 0

3 years ago

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deon is running at a velocity of 7.9 meters per second. deon has a total mass of 95 kilograms including his helmet, uniform, pad

brilliants [131]

Answer:

712.5 kg m/s

Explanation:

Work out the total momentum before the event (before the collision):

p = m × v

Massof Deon = 95kg

Velocity =7.5m/s

Mass of Chuck = 120kg

Velocity = 0m/s

Momentum of Deon before = 95 × 7.5 = 712.5 kg m/s

Momentum of Chuck before = 120 × 0 = 0 kg m/s

Total momentum before = 712.5+ 0 = 712.5 kg m/s

Working out the total momentum after the event (after the collision):

Because momentum is conserved, total momentum afterwards = 712.5 kg m/s

6 0

4 years ago

Pleaseeeee help me with b, c, and d. There are no angles.

taurus [48]

Answer:

a. 150 J

b. 150 J

c. 0 J

d. 0 J

Explanation:

The given parameters are;

The horizontal force with which the man pulls the canister, F = 50 N

The distance he moves the vacuum cleaner, d = 3.0 m

a. Work done, W = Force applied, F × Distance moved by the force, d

Therefore, for the work done by the 50 N force on the canister, we have;

W = 50 N × 3.0 m = 150 N·m = 150 J

b. Given that he pulls the canister at a constant speed, we have;

The acceleration of the canister, a = 0 m/s²

Therefore, the net force on the canister, $F_{NET}$ = F - $F_{Friction}$ = m × a

Where;

m = The mass of the canister

a = The acceleration of the canister

F = The applied force = 50 N

$F_{Friction}$ = The force of friction

∴ $F_{NET}$ = m × a = m × 0 m/s² = 0 N

Therefore;

$F_{NET}$ = F - $F_{Friction}$ = 0 N

From which we have;

F = $F_{Friction}$ = 50 N (The applied force, F is equal to the force of friction,

The work done by friction = The force of friction × The distance in which the force of friction acts

∴ The work done by friction = $F_{Friction}$ × d - 50 N × 3.0 m = 150 J

The work done by friction = 150 J

c. The normal force, N acts perpendicular to the force of friction

The distance the canister moves in the perpendicular direction, $d_p$ = 0 m

∴ The work done by the normal direction = N × $d_p$ = N × 0 m = 0 J

The work done by the normal direction = 0 J

d. The vacuum weight, W, acts on the same line as the normal force but in the opposite direction to the normal force, N

Therefore, the weight, W, acts perpendicular to the line of motion of canister

The distance the canister moves in the direction of the weight, $d_{wieght}$ = 0 m

Therefore, the work done by the weight = W × $d_{wieght}$ = W × 0 m = 0 J

The work done by the weight = 0 J

7 0

3 years ago

Assume that in 2010 the United States will need 2.0×1012 watts of electric power produced by thousands of 1000 MW power plants.

Alex73 [517]

Answer:

1752.14 tonnes per year.

Explanation:

To solve this exercise it is necessary to apply the concepts related to power consumption and power production.

By conservation of energy we know that:

$\dot{P} = \bar{P}$

Where,

$\dot{P} =$ Production of Power

$\bar{P} =$ Consumption of power

Where the production of power would be,

$\dot{P} = m \dot{E}\eta$

Where,

m = Total mass required

$\dot{E} =$ Energy per Kilogram

$\eta =$ Efficiency

The problem gives us the aforementioned values under a production efficiency of 45%, that is,

$\dot{P} = \bar{P}$

$m \dot{E}\eta = \bar{P}$

Replacing the values we have,

$m(8*10^13)(0.45) = 2*10^{12}$

Solving for m,

$m = \frac{ 2*10^{12}}{(8*10^13)(0.45)}$

$m = 0.0556 \frac{kg}{s}$

We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

$m = 0.556\frac{kg}{s}*(\frac{3.1536*10^7s}{1year})(\frac{1ton}{1000kg})$

$m = 1752.14$ tonnes per year.

8 0

3 years ago

A rocket travels 1.3 km in 62 ms. What is its average speed in m⋅s−1? Do not give your answer in scientific notation. The answer

hjlf

Answer:

Average speed = 0.35 m/s

Explanation:

Given the following data;

Distance = 1.3 Km

Time = 62 minutes

To find the average speed in m/s;

First of all, we would convert the quantities to their standard unit (S.I) of measurement;

Conversion:

1.3 kilometres to meters = 1.3 * 1000 = 1300 meters

For time;

1 minute = 60 seconds

62 minutes = X

Cross-multiplying, we have;

X = 62 * 60

X = 3720 seconds

Now, we can calculate the average speed in m/s using the formula;

$Speed = \frac {distance}{time}$

$Speed = \frac {1300}{3720}$

Average speed = 0.35 m/s

7 0

4 years ago