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4 years ago

How far below an initial straight-line path will a projectile fall in one second

Physics

1 answer:

Anni [7]4 years ago

3 0

Answer:

9.8 m

Explanation:

The rate of free fall of a body does not depend on its mass or shape in the absence of air resistance. This property of gravitational force is proved by the famous leaning tower of Pisa experiment by Galileo.

The gravitational force is responsible for the gravitational acceleration of a body.

When considering the gravitational acceleration near the surface of Earth it has the constant value g = 9.8 m/s irrespective of the mass of the body.

This phenomenon states that every projectile falls at the same rate such that its velocity keeps on increasing to maintain the constancy of g.

So, when a projectile fall initially from a certain height for one second, it would have traveled only 9.8 m.

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If the potential is given by v = xy - 3z-2, then the electric field has a y-component of?

kiruha [24]

If the potential is given by v = xy - 3z-2, then the electric field has a y-component of X

When the charge is present in any form, a point in space has an electric field that is connected to it. The value of E, often known as the electric field strength, electric field intensity, or just the electric field, expresses the strength and direction of the electric field.

Each location in space where a charge exists in any form can be considered to have an electric field attached to it. The electric force per unit charge is another name for an electric field. The electric field's equation is given as E = F / Q. Volts per meter (V/m) is the electric field's SI unit. Newton's per coulomb unit is the same as this one.

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3 0

2 years ago

Which part of the electromagnetic spectrum have the highest level of photon energy

Pani-rosa [81]

The energy carried by one photon is directly proportional to its
frequency. So the photon energy is greatest for the electromagnetic
waves with the highest frequency / shortest wavelengths.

That's why when you get past visible light and on up through ultraviolet,
X-rays, and gamma rays, the radiation becomes dangerous ==> each
photon carries enough energy to tear electrons away from their atoms,
ripping molecules apart and damaging cells.

The photon with the highest energy is a gamma-ray photon.

4 0

3 years ago

A ball is thrown toward a cliff of height h with a speed of 33 m/s and an angle of 60∘ above horizontal. It lands on the edge of

Veseljchak [2.6K]

Answer:

(a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

Explanation:

Given that,

Speed = 33 m/s

Angle = 60°

Time = 3.0 sec

(a). We need to calculate the height of the cliff

Using equation of motion

$h=ut-\dfrac{1}{2}gt^2$

$h=(u\sin60)\times t-\dfrac{1}{2}gt^2$

Put the value into the formula

$h=33\times\sin60\times3.0-\dfrac{1}{2}\times9.8\times(3.0)^2$

$h=41.6\ m$

(b). We need to calculate the maximum height of the ball

Using formula of height

$h_{max}=\dfrac{(u\sin\theta)^2}{2g}$

Put the value into the formula

$h=\dfrac{(33\sin60)^2}{2\times 9.8}$

$h=41.67\ m$

(c). We need to calculate the vertical component of velocity of ball

Using equation of motion

$v=u-gt$

$v=u\sin\theta-gt$

Put the value into the formula

$v_{y}=33\times\sin 60-9.8\times3.0$

$v_{y}=-0.82\ m/s$

We need to calculate the horizontal component of velocity of ball

Using formula of velocity

$v_{x}=u\cos\theta$

Put the value into the formula

$v_{x}=33\times\cos60$

$v_{x}=16.5\ m/s$

We need to calculate the ball's impact speed

Using formula of velocity

$v=\sqrt{v_{x}^2+v_{y}^2}$

Put the value into the formula

$v=\sqrt{(16.5)^2+(-0.82)^2}$

$v=16.52\ m/s$

Hence, (a). The height of the cliff is 41.67 m.

(b). The maximum height of the ball is 41.67 m

(c). The ball's impact speed is 16.52 m/s.

3 0

3 years ago

Two objects, one with a mass of and the other with a force of 30.0kg experience a gravitational force of attraction of 7.50 * 10

nydimaria [60]

Answer:

Explanation:

The formula for this is

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the gravitational constant, m1 is the mass of one object and m2 is the mass of the other object. We are looking for r, the distance between the centers of their masses.