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2 years ago

It opens or close the circuit

Physics

1 answer:

gayaneshka [121]2 years ago

3 0

Answer:

The person above me is right i had a test a couple of days ago and thats kinda what u put and got it right!

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An object moving with a speed or 67m/s and has kinetic energy of 500 J what is the mass of the object

saveliy_v [14]

To solve this equation, simply plug the values into the equation for calculating kinetic energy.

KE = 1/2mv^2
500 = 1/2(m)(67^2)
500 =2244.5m
m = 500/2244.5 = 0.222 kg.

8 0

2 years ago

In the standing waves experiment, the string has a mass of 31.2 g and a length of 0.7 m. The string is connected to a mechanical

mestny [16]

Answer:

linear density of the string = 4.46 × 10⁻⁴ kg/m

Explanation:

given,

mass of the string = 31.2 g

length of string = 0.7 m

linear density of the string = $\dfrac{mass\ of\ string}{length}$

linear density of the string = $\dfrac{31.2\times 10^{-3}\ kg}{0.7\ m}$

linear density of the string = 44.57 × 10⁻³ kg/m

linear density of the string = 4.46 × 10⁻⁴ kg/m

7 0

3 years ago

If the second harmonic of a certain string is 42 Hz, what is the fundamental frequency of the string?

stich3 [128]

I would think that you would have to do 42/2=21Hz, but I'm not sure...

5 0

2 years ago

A vector has an x-component

NeTakaya

Answer:

34.45m

Explanation:

Magnitude of a vector is equal to the square root of sum of squares of x & y vectors.

Magnitude = $\sqrt({x^2} + {y^2})$

= $\sqrt({19.5^2} + {28.4^2})$

=34.45m

5 0

2 years ago

A car accelerates in the +x direction from rest with a constant acceleration of a1 = 1.76 m/s2 for t1 = 20 s. At that point the

alex41 [277]

Answer:

(a) $v_1 = a_1t_1 = 1.76 t_1$

(b) It won't hit

Explanation:

(a) the car velocity is the initial velocity (at rest so 0) plus product of acceleration and time t1

$v_1 = v_0 + a_1t_1 = 0 +1.76t_1 = 1.76t_1$

(b) The velocity of the car before the driver begins braking is

$v_1 = 1.76*20 = 35.2m/s$

The driver brakes hard and come to rest for t2 = 5s. This means the deceleration of the driver during braking process is

$a_2 = \frac{\Delta v_2}{\Delta t_2} = \frac{v_2 - v_1}{t_2} = \frac{0 - 35.2}{5} = -7.04 m/s^2$

We can use the following equation of motion to calculate how far the car has travel since braking to stop

$s_2 = v_1t_2 + a_2t_2^2/2$

$s_2 = 35.2*5 - 7.04*5^2/2 = 88 m$

Also the distance from start to where the driver starts braking is

$s_1 = a_1t_1^2/2 = 1.76*20^2/2 = 352$

So the total distance from rest to stop is 352 + 88 = 440 m < 550 m so the car won't hit the limb

8 0

2 years ago

It opens or close the circuit​

It opens or close the circuit