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Afina-wow [57]
3 years ago
7

a diver on a board 7.5m above the water walks off the end with a horizontal velocity of 2.3 m/s when they hit the surface of the

water how far away from the edge of the board will they land ?
Physics
1 answer:
m_a_m_a [10]3 years ago
6 0

Answer:

2.846m

Explanation:

The diver is performing projectile motion.

To find x(final), we are going to use the equation x(final) = v(initial)*t + x(initial)

x(initial) = 0

x(final) = ?

v(initial) = 2.3 m/s

we don't know t

To find t we will use y(final) = 1/2*(-9.8)*t^2 + v(initial in the y dir.)*t + y(initial)

- 9.8 in the acceleration in the y dir.

y(final) = 0

y(initial) = 7.5

v(initial in the y dir.) = 0

If we solve for t we get: t = 1.237s

Now we have all the components to solve for x(final) in x(final) = v(initial)*t + x(initial)

x(final) = 2.3*1.237 + 0

x(final) = 2.846m

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How do mechanical waves compare with electromagnetic waves?
Naddika [18.5K]

Answer:

Main Difference Between Mechanical and Electromagnetic waves

A wave is composed of some kind of disturbance that propagates. We can classify waves into many different types based on their properties. One of the properties of the waves depends on whether they need a medium to propagate or not. The primary difference between electromagnetic and mechanical waves is also based on this property. Mechanical waves need a medium, while electromagnetic waves do not need a medium to propagate. Electromagnetic waves can travel through a vacuum. The other differences between mechanical and electromagnetic waves are given below:

Electromagnetic waves can travel through a vacuum, that is an empty space, whereas mechanical waves cannot. They need a medium to travel such as water or air. Ripples in a pond are an example of mechanical waves whereas electromagnetic waves include light and radio signals, which can travel through the vacuum of space.

Mechanical waves can be classed as elastic waves because their transmission depends on the medium's (water, air etc.) elastic properties.

Electromagnetic waves are caused because of the varying magnetic and electric fields. They are produced by the vibration of the charged particles.

Because of these differences, the speed of each type of wave varies significantly. Electromagnetic waves travel at the speed of light but mechanical waves are far slower.

                                                                            Have a wonderful day!

                                                                                                ~Lillith of brainly~

6 0
2 years ago
A current of 5.0 a flows through an electrical device for 10 seconds. how many electrons flow through this device during this ti
melisa1 [442]
1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec

Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s

In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons

4 0
3 years ago
A soft tennis ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.10 m. (a) Calculate its ve
Gemiola [76]

Answer:

(a)   v = 5.42m/s

(b)   vo = 4.64m/s

(c)   a = 2874.28m/s^2

(d)   Δy = 5.11*10^-3m

Explanation:

(a) The velocity of the ball before it hits the floor is given by:

v=\sqrt{2gh}        (1)

g: gravitational acceleration = 9.8m/s^2

h: height where the ball falls down = 1.50m

v=\sqrt{2(9.8m/s^2)(1.50m)}=5.42\frac{m}{s}

The speed of the ball is 5.42m/s

(b) To calculate the velocity of the ball, after it leaves the floor, you use the information of the maximum height reached by the ball after it leaves the floor.

You use the following formula:

h_{max}=\frac{v_o^2}{2g}       (2)

vo: velocity of the ball where it starts its motion upward

You solve for vo and replace the values of the parameters:

v_o=\sqrt{2gh_{max}}=\sqrt{2(9.8m/s^2)(1.10m)}=4.64\frac{m}{s}

The velocity of the ball is 4.64m/s

(c) The acceleration is given by:

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-(-5.42m)/s}{3.50*10^{-3}s}=2874.285\frac{m}{s^2}

a=\frac{\Delta v}{\Delta t}=\frac{v_o-v}{3.50*10^{-3}s}=\frac{4.64m/s-5.42m/s}{3.50*10^{-3}s}=-222.85\frac{m}{s^2}

The acceleration of the ball is 2874.28/s^2

(d) The compression of the ball is:

\Deta y=\frac{v^2}{2(a)}=\frac{(5.42m/s)^2}{2(2874.28m/s^2)}=5.11*10^{-3}m

THe compression of the ball when it strikes the floor is 5.11*10^-3m

4 0
3 years ago
A ball is tossed with enough speed straight up so that it is in the air several seconds. (a) What is the velocity of the ball wh
irina1246 [14]

(a) Zero

When the ball reaches its highest point, the direction of motion of the ball reverses (from upward to downward). This means that the velocity is changing sign: this also means that at that moment, the velocity must be zero.

This can be also understood in terms of conservation of energy: when the ball is tossed up, initially it has kinetic energy

K=\frac{1}{2}mv^2

where m is the ball's mass and v is the initial speed. As it goes up, this kinetic energy is converted into potential energy, and when the ball reaches the highest point, all the kinetic energy has been converted into potential energy:

U=mgh

where g is the gravitational acceleration and h is the height of the ball at highest point. At that point, therefore, the potential energy is maximum, while the kinetic energy is zero, and so the velocity is also zero.

(b) 9.8 m/s upward

We can find the velocity of the ball 1 s before reaching its highest point by using the equation:

a=\frac{v-u}{t}

where

a = g = -9.8 m/s^2 is the acceleration due to gravity, which is negative since it points downward

v = 0 is the final velocity (at the highest point)

u is the initial velocity

t = 1 s is the time interval

Solving for u, we find

u=v-at = 0 -(-9.8 m/s^2)(1 s)= +9.8 m/s

and the positive sign means it points upward.

(c) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where

v = 0 is the final velocity (at the highest point)

u = 9.8 m/s is the initial velocity

Substituting, we find

\Delta v = 0 - (+9.8 m/s)=-9.8 m/s

(d) 9.8 m/s downward

We can find the velocity of the ball 1 s after reaching its highest point by using again the equation:

a=\frac{v-u}{t}

where this time we have

a = g = -9.8 m/s^2 is the acceleration due to gravity, still negative

v  is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

t = 1 s is the time interval

Solving for v, we find

v = u+at = 0 +(-9.8 m/s^2)(1 s)= -9.8 m/s

and the negative sign means it points downward.

(e) -9.8 m/s

The change in velocity during the 1-s interval is given by

\Delta v = v -u

where here we have

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = 0 is the initial velocity (at the highest point)

Substituting, we find

\Delta v = -9.8 m/s - 0=-9.8 m/s

(f) -19.6 m/s

The change in velocity during the overall 2-s interval is given by

\Delta v = v -u

where in this case we have:

v = -9.8 m/s is the final velocity (1 s after reaching the highest point)

u = +9.8 m/s is the initial velocity (1 s before reaching the highest point)

Substituting, we find

\Delta v = -9.8 m/s - (+9.8 m/s)=-19.6 m/s

(g) -9.8 m/s^2

There is always one force acting on the ball during the motion: the force of gravity, which is given by

F=mg

where

m is the mass of the ball

g = -9.8 m/s^2 is the acceleration due to gravity

According to Newton's second law, the resultant of the forces acting on the body is equal to the product of mass and acceleration (a), so

mg = ma

which means that the acceleration is

a= g = -9.8 m/s^2

and the negative sign means it points downward.

7 0
3 years ago
How is the mass of a body measured ? Write in brief.<br>​
skad [1K]

Answer:

Mass is the quantity of matter contained in a body. The mass of an object is measured in kilogram (kg).

Explanation:

3 0
3 years ago
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