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joja [24]
3 years ago
12

What is centripetal forcehello​

Physics
2 answers:
Gwar [14]3 years ago
8 0

Answer:

A centripetal force is a net force that acts on an object to keep it moving along a circular path

Bond [772]3 years ago
5 0

Answer:

A force that acts on a body moving in a circular path and directed towards the centre around which the body is moving is called Centripetal force.

Explanation:

When an object travels around a circular path with a constant speed, it experiences an accelerating centripetal force towards the centre.

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When trying to predict the lowest temperature that will be reached overnight, forecasters pay close attention to the dew-point t
wariber [46]
<h3>Answer:</h3>

"<u>This is especially true for dewpoints above 55 degrees. If the dewpoint happens to be much lower than the temperature, the air will cool off much more rapidly at night than if the dewpoint was closer to the temperature in the evening</u>."

7 0
3 years ago
What makes a planet different from other celestial bodies?.
Serggg [28]

Answer:

It's more habitable.

Explanation:

The atmosphere, calculated to equations, are a lot more pulled down.

5 0
3 years ago
A car travels at 40 mph. what distance will it travel in 3 hours?​
lukranit [14]

Car speed = 40m/ hour

3 hours = 40 × 3 = 120m /hour

8 0
3 years ago
Read 2 more answers
How do I go about this?
Anna71 [15]

Hi there!

(a)

Recall that:
W = F \cdot d = Fdcos\theta

W = Work (J)
F = Force (N)
d = Displacement (m)

Since this is a dot product, we only use the component of force that is IN the direction of the displacement. We can use the horizontal component of the given force to solve for the work.

W =248(56)cos(30) = 12027.36 J

To the nearest multiple of ten:
W_A = \boxed{12030 J}

(b)
The object is not being displaced vertically. Since the displacement (horizontal) is perpendicular to the force of gravity (vertical), cos(90°) = 0, and there is NO work done by gravity.

Thus:
\boxed{W_g = 0 J}

(c)
Similarly, the normal force is perpendicular to the displacement, so:
\boxed{W_N = 0 J}

(d)

Recall that the force of kinetic friction is given by:
F_{f} =\mu_k mg

Since the force of friction resists the applied force (assigned the positive direction), the work due to friction is NEGATIVE because energy is being LOST. Thus:
W_f = -\mu_k mgd\\W_f = - (0.1)(56)(9.8)(56) = -3073.28 J

In multiples of ten:
\boxed{W_f = -3070 J}

(e)
Simply add up the above values of work to find the net work.

W_{net} = W_A + W_f \\\\W_{net} = 12027.36 + (-3073.28) = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}}

(f)
Similarly, we can use a summation of forces in the HORIZONTAL direction. (cosine of the applied force)
F_{net} = F_{Ax} - F_f

W = F_{net} \cdot d = (F_{Ax} - F_f)

W = (F_Acos(30) - \mu_k mg)d\\W = (248cos(30) - 0.1(56)(9.8)) * 56 \\\\W = 8954.08 J

Nearest multiple of ten:
\boxed{W_{net} = 8950 J}

5 0
2 years ago
A garden hose having with an internal diameter of 1.1 cm is connected to a (stationary) lawn sprinkler that consists merely of a
trapecia [35]

Answer:

Water leaves the sprinkler at a speed of 2.322 m/sec

Explanation:

We have given internal diameter of the garden d_1=1.1cm

Speed of water in the hose is v_1=0.95m/sec

Number of holes n = 22

Diameter of each holes d_2=15cm

According to continuity equation A_1v_1=A_2v_2

d_1^2\times v_1=22\times d_2^2v_2

1.1^2\times 0.95=22\times 0.15^2\times v_2

v_2=2.322m/sec

So water leaves the sprinkler at a speed of 2.322 m/sec

6 0
3 years ago
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