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3 years ago

What is an object’s acceleration if it is moving at 30 m/s and comes to a stop in 5 s? –30 m/s2 –6 m/s2 30 m/s2 6 m/s2

Physics

2 answers:

Readme [11.4K]3 years ago

7 0

Answer:

The acceleration is -6 m/s2

Explanation:

The equation for the speed of a body is

V = Vo + a t

Where

Vo is the initial velocity, a is the acceleration and t is the time,

When the body stops the speed is zero

0 = Vo + a t

a = Vo / t

substitute values

a = - 30/5

a = - 6 m/s2

The negative sign indicates that acceleration opposite movement

Tanya [424]3 years ago

3 0

Answer:32m/s

Explanation:

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All electric devices are required to have identifying plates that specify their electrical characteristics. The plate on a certa

masha68 [24]

Answer:

R= 20 ohm

Explanation:

Given that

Current ,I = 6 A

Voltage difference ,ΔV = 120 V

Lets take resistance of the stem iron = R

We know that ,the relationship between current ,voltage difference and resistance is given as

ΔV = I R

$R=\dfrac{\Delta V}{I}$

Now by putting the values in the above equation we get

$R=\dfrac{120}{6}\ ohm$

R= 20 ohm

Therefore the resistance of the steam will be 20 ohm.

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3 years ago

How much power is needed to lift the 200-N object to a height of 4 m in 4 s?

Vitek1552 [10]

Answer: 2000 watts

Explanation:

Given that,

power = ?

Weight of object = 200-N

height = 4 m

Time = 4 s

Power is the rate of work done per unit time i.e Power is simply obtained by dividing work by time. Its unit is watts.

i.e Power = work / time

(since work = force x distance, and weight is the force acting on the object due to gravity)

Then, Power = (weight x distance) / time

Power = (200N x 4m) / 4s

Power = 8000Nm / 4s

Power = 2000 watts

Thus, 2000 watts of power is needed to lift the object.

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3 years ago

In the model of the hydrogen atom due to Niels Bohr, the electron moves around the proton at a speed of 3.3 × 106 m/s in a circl

irga5000 [103]

Answer:

$1.5048\times 10^{-23}\ Am^2$

Explanation:

q = Charge of proton = $1.6\times 10^{-19}\ C$

r = Radius of circle = $5.7\times 10^{-11}\ m$

v = Velocity of proton = $3.3\times 10^6\ m/s$

Magnetic moment is given by

$M=\frac{1}{2}qrv\\\Rightarrow M=\frac{1}{2}1.6\times 10^{-19}\times 5.7\times 10^{-11}\times 3.3\times 10^6\\\Rightarrow M=1.5048\times 10^{-23}\ Am^2$

The magnetic moment associated with this motion is $1.5048\times 10^{-23}\ Am^2$

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Olegator [25]

Answer:

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Naya [18.7K]

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