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3 years ago

How are speed and velocity similar ?

Physics

2 answers:

Anna [14]3 years ago

8 0

I would say A to Bc that what I think it is

Serhud [2]3 years ago

3 0

I agree with you the answer is A

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What happens when you drink water that was in the car heated then you drink it. Make sure to explain your answer to make you as

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When you drink water that was in the car heated, you can get headaches and it would make you dizzy. I know a little bit because it happen to my dad.

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4 years ago

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At a certain location, Earth has a magnetic field of 0.60 ✕ 10−4 T, pointing 75° below the horizontal in a north-south plane. A

saveliy_v [14]

Answer with Explanation:

We are given that

Magnetic field,B= $0.6\times 10^{-4} T$

$\theta=75^{\circ}$

Length of wire,l=15 m

Current,I=19 A

a.We have to find the magnitude of magnetic force and direction of magnetic force.

Magnetic force,F= $IBlsin\theta$

Using the formula

$F=0.6\times 10^{-4}\times 15\times 19sin75$

$F=16.5\times 10^{-3} N$

Direction= $tan\theta=cot(90-75)=tan15^{\circ}$

$\theta=15^{\circ}$

15 degree above the horizontal in the northward direction.

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3 years ago

A police car in hot pursuit goes speeding past you. While the siren is approaching, the frequency of the sound you hear is 5500

larisa [96]

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3 years ago

Write the untis of following physical quantities electric current , force, frequency, density

Shtirlitz [24]

Answer:

1) electric current - ampere

2) force - joule

3 frequency - hertz

4) density - kilogram per cubic metre

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3 years ago

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a man drags a 8.10 kg bag of mulch at a constant speed, applying a 29.5 N at 38°. what is the coefficient of friction?

lesantik [10]

Answer:

The coefficient of friction is 0.38.

Explanation:

The free body diagram is drawn below.

Let $f$ be frictional force acting in the backward direction as shown. Let the coefficient of friction be $\mu$ . Let $N$ be the normal reaction force acting on the bag.

Given:

Mass of the bag is, $m=8.10\textrm{ kg}$

Force acting at $\theta = 38$ ° is $F= 29.5\textrm{ N}$

Acceleration due to gravity is, $g=9.8\textrm{ }m/s^{2}$

The force F can be resolved into its components as $F_{x}=F \cos \theta$ and $F_{y}=F \sin \theta$

Therefore,

$F_{x}=29.5\cos(38)=23.25\textrm{ N}\\F_{y}=29.5\sin(38)=18.16\textrm{ N}$

Now, as there is no acceleration in vertical direction, therefore,

Sum of upward forces = Sum of downward forces

$N+F_{y}=mg\\N=mg-F_{y}=8.10\times 9.8-18.16\\N=79.38-18.16=61.22\textrm{ N}$

Now, as the bag is moving at a constant speed, so acceleration in the horizontal direction is also zero as acceleration is the rate of change of velocity.

Therefore, backward force = forward force.

$f=F_{x}\\f=23.25\textrm{ N}$

Now, frictional force is given as:

$f=\mu N\\\mu = \frac{f}{N}=\frac{23.25}{61.22}=0.38$

Therefore, the coefficient of friction is 0.38.

8 0

4 years ago