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Rzqust [24]
2 years ago
7

The illustration below shows a ball on a rope. The mass of the ball is 0.4 kg. Assume that the mass of the rope is negligible. A

ssume that the friction is negligible. What is the maximum speed of the ball as it swings back and forth?
A)0.14 m/s
B)21 m/s
C)7.0 m/s
D)49 m/s
Physics
1 answer:
Dafna11 [192]2 years ago
7 0

Answer:

A

Explanation: Wave length

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Please help me, it's a simple equation
castortr0y [4]

Because the elevator moves at a constant speed, it's in equilibrium and the net force acting on it is zero. Then the tension in the cable exactly equals the magnitude of the elevator's weight, which is

(3000 kg) (9.80 m/s²) = 29,400 N

3 0
2 years ago
You pull a block of mass m across across a frictionless table with a constant force. you also pull with an equal constant force
KiRa [710]
For any mass m:

a = F/m
v = √2*F/m*s = √2F/sm = k/√m
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SO
Both will have same energy
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4 0
3 years ago
Zinc can not be changed into a simpler substance by a chemical or physical process. Therefore, zinc is classified as:
bonufazy [111]
The correct answer I think is a compound
7 0
3 years ago
A boat of mass 250 kg is coasting, with its engine in neutral, through the water at speed 1.00 m/s when it starts to rain with i
Vera_Pavlovna [14]

Answer:

Explanation:

1. We use the conservation of momentum for before the raining and after. And also we take into account that in 0.5h the accumulated water is

100kg/h*0.5h = 50kg

p_{b}=p_{a}\\M_{b}v_{b}=(M_{b}+m_{r})v_{a}\\v_{a}=\frac{M_{b}v_{b}}{M_{b}+m_{r}}=\frac{250kg*1\frac{m}{s}}{250kg+50kg}=0.83\frac{m}{s}

2. the momentum does not conserve because the drag force of water makes that the boat loses velocity

3. If we assume that the force of the boat before the raining is

F=ma=m\frac{v-v_{0}}{t-t_{0}}=250kg\frac{1m}{s^{2}}=250N

where we have assumed that the acceleration of the boat is 1m/s{2} just before the rain starts

And if we take the net force as

F_{net}=M_{b}a_{net}=F-F_{d}=250N-bv^{2}\\F_{net}=250N-0.5N\frac{s^{2}}{m^{2}}(1\frac{m}{s})^{2}=249.5N\\a_{net}=\frac{249.5N}{M_{b}}=\frac{249.5N}{250kg}=0.99\frac{m}{s^{2}}

where we take v=1m/s because we are taking into account tha velocity just after the rain stars.

I hope this is useful for you

regards

5 0
3 years ago
A 100 Newton force applied to a machine lifts a 400 N object. What is the actual mechanical advantage of
uranmaximum [27]

Answer:

m = 4

Explanation:

It is given that,

A 100 Newton force applied to a machine lifts a 400 N object.  

We need to find the mechanical advantage of  this machine.

The mechanical advantage of a machine is given by the ratio of output force to the input force.

Here, output force is 400 N and input force is 100 N

So, mechanical advantage becomes :

m=\dfrac{400}{100}\\\\m=4

So, the mechanical advantage of this machine is 4.

7 0
3 years ago
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