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charle [14.2K]
3 years ago
5

Long ding dong what it do?

Physics
2 answers:
Nezavi [6.7K]3 years ago
7 0

Answer:

long very very

Andre45 [30]3 years ago
3 0

Answer:

nun wbu

Explanation:

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you find out that it takes 2 sec. for the swing to complete one cycle. what is the swing`s period and frequency?
Gnesinka [82]

The time it takes "to complete one cycle" is a perfect definition for <em>period</em>.
So the <em>2 sec</em> that you measured is the swing's period.

Frequency is just   1/period  (the 'reciprocal' of the period).
For this swing, the frequency is   1/(2sec)  =  0.5 per second = <em>0.5 Hz</em>.

( "Hertz" means "per second")


6 0
3 years ago
How many aluminum atoms are in the following compound?<br> 4 Al(OH)3
vovikov84 [41]

Answer:

There are 4 Aluminum atoms in 4Al(OH)3.

3 0
3 years ago
I dont understand ..............
ivann1987 [24]

Take the upward and to-the-right directions to be positive (so down and to-the-left are negative).

The vertical forces acting on the object cancel, 6 N - 6 N = 0.

The horizontal forces exert a net force of 20 N - 3 N = 17 N. This net force is positive, so it points to the right. So the answer is A.

7 0
3 years ago
A 99.5 N grocery cart is pushed 12.9 m along an aisle by a shopper who exerts a constant horizontal force of 34.6 N. The acceler
Romashka [77]

1) 9.4 m/s

First of all, we can calculate the work done by the horizontal force, given by

W = Fd

where

F = 34.6 N is the magnitude of the force

d = 12.9 m is the displacement of the cart

Solving ,

W = (34.6 N)(12.9 m) = 446.3 J

According to the work-energy theorem, this is also equal to the kinetic energy gained by the cart:

W=K_f - K_i

Since the cart was initially at rest, K_i = 0, so

W=K_f = \frac{1}{2}mv^2 (1)

where

m is the of the cart

v is the final speed

The mass of the cart can be found starting from its weight, F_g = 99.5 N:

m=\frac{F_g}{g}=\frac{99.5 N}{9.8 m/s^2}=10.2 kg

So solving eq.(1) for v, we find the final speed of the cart:

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(446.3 J)}{10.2 kg}}=9.4 m/s

2) 2.51\cdot 10^7 J

The work done on the train is given by

W = Fd

where

F is the magnitude of the force

d is the displacement of the train

In this problem,

F=4.28 \cdot 10^5 N

d=586 m

So the work done is

W=(4.28\cdot 10^5 N)(586 m)=2.51\cdot 10^7 J

3)  2.51\cdot 10^7 J

According to the work-energy theorem, the change in kinetic energy of the train is equal to the work done on it:

W=\Delta K = K_f - K_i

where

W is the work done

\Delta K is the change in kinetic energy

Therefore, the change in kinetic energy is

\Delta K = W = 2.51\cdot 10^7 J

4) 37.2 m/s

According to the work-energy theorem,

W=\Delta K = K_f - K_i

where

K_f is the final kinetic energy of the train

K_i = 0 is the initial kinetic energy of the train, which is zero since the train started from rest

Re-writing the equation,

W=K_f = \frac{1}{2}mv^2

where

m = 36300 kg is the mass of the train

v is the final speed of the train

Solving for v, we find

v=\sqrt{\frac{2W}{m}}=\sqrt{\frac{2(2.51\cdot 10^7 J)}{36300 kg}}=37.2 m/s

7 0
4 years ago
When jumping down from a table to the ground, it is helpful to bend your knees when you hit the ground because...
forsale [732]

Answer:

This reduces the average force applied during the landing process/ or you can say it reduces the impact your body takes.

Explanation:

7 0
3 years ago
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