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2 years ago

HELPPPP I HAVE A TEST TODAY AND I LITERALLY CAN'T WITH THIS

Physics

1 answer:

dsp732 years ago

7 0

The horizontal component of velocity is

(22 m/s) • cosine(62°).

The vertical component of velocity is

(22 m/s) • sine(62°).

These are the original components, right after the kick. As time goes on, the horizontal one doesn't change. But the vertical one gets bigger and bigger, because gravity is accelerating the ball downward.

That's the complete story of projectile motion.

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A lightweight string is wrapped several times around the rim of a small hoop. If the free end of the string is held in place and

MariettaO [177]

Answer:

Explanation:

Let T be the tension

For linear motion of hoop downwards

mg -T = ma , m is mass of the hoop . a is linear acceleration of CG of hoop .

For rotational motion of hoop

Torque by tension

T x R , R is radius of hoop.

Angular acceleration be α,

Linear acceleration a = α R

So TR = I α

= I a / R

a = TR² / I

Putting this value in earlier relation

mg -T = m TR² / I

mg = T ( 1 + m R² / I )

T = mg / ( 1 + m R² / I )

mg / ( 1 + R² / k² )

Tension is less than mg or weight because denominator of the expression is more than 1.

5 0

3 years ago

What is the gravitational potential energy of a 2.5kg object that is 300m above the surface of the earth? g=10m/s

Alexus [3.1K]

Answer:

7350 J

Explanation:

Gravitational Potential Energy: This is defined as the energy possessed by a body due to it's position in the gravitational field. The S.I unit is Joules(J).

Applying,

E.p = mgh..................... Equation 1

Where E.p = Gravitational potential Energy, m = mass of the object, h = height of the object above the surface of the earth, g = acceleration due to gravity.

Given: m = 2.5 kg, h = 300 m

Constant: g = 9.8 m/s²

Substitute these values into equation 1

E.p = 2.5(300)(9.8)

E.p = 7350 J.

4 0

3 years ago

What is the condition to earths atmosphere at a givin time and pace

alina1380 [7]

The condition of earths atmosphere at a given time and place is the weather.

6 0

3 years ago

Two thin concentric spherical shells of radii r1 and r2 (r1 < r2) contain uniform surface charge densities V1 and V2, respect

Lyrx [107]

Answer:

Answer is explained in the explanation section below.

Explanation:

Solution:

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

So,

a) 0 < r < r1 :

We know that the Electric field inside the thin hollow shell is zero, if there is no charge inside it.

Hence, E = 0 for r < r1

b) r1 < r < r2:

Electric field =?

Let, us consider the Gaussian Surface,

E x 4 $\pi$ $r^{2}$ = $\frac{Q1}{E_{0} }$

So,

Rearranging the above equation to get Electric field, we will get:

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$

Multiply and divide by $r1^{2}$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ x $\frac{r1^{2} }{r1^{2} }$

Rearranging the above equation, we will get Electric Field for r1 < r < r2:

E= (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

c) r > r2 :

Electric Field = ?

E x 4 $\pi$ $r^{2}$ = $\frac{Q1 + Q2}{E_{0} }$

Rearranging the above equation for E:

E = $\frac{Q1+Q2}{E_{0} . 4 \pi. r^{2} }$

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

As we know from above, that:

$\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Then, Similarly,

$\frac{Q2}{E_{0} . 4 \pi. r^{2} }$ = (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

So,

E = $\frac{Q1}{E_{0} . 4 \pi. r^{2} }$ + $\frac{Q2}{E_{0} . 4 \pi. r^{2} }$

Replacing the above equations to get E:

E = (σ1 x $r1^{2}$ ) /( $E_{0}$ x $r^{2}$ ) + (σ2 x $r2^{2}$ ) /( $E_{0}$ x $r^{2}$ )

Now, for

d) Under what conditions, E = 0, for r > r2?

For r > r2, E =0 if

σ1 x $r1^{2}$ = - σ2 x $r2^{2}$

4 0

3 years ago

In a compression wave, particles in the medium move

nikdorinn [45]

in the same direction as the wave

Explanation:

In a compression wave, the particles in the medium moves in the same direction as the wave source.

A wave is generally defined as a disturbance that transmits energy.

There are two types of waves based on the direction through which they are propagated.
Transverse waves are directed perpendicularly in the direction of propagation.
Examples are electromagnetic waves.
Longitudinal waves are parallel to their source. Examples are sound waves, p-waves.
They are made up of series of rarefaction and compression.

learn more:

Waves brainly.com/question/3183125

#learnwithBrainly

8 0

2 years ago