Ask question

Ask question

All categories

LUCKY_DIMON [66]

3 years ago

5

PLZ help !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

1 answer:

vlada-n [284]3 years ago

6 0

Answer:

The leftmost red square.

Explanation:

As the earth surface we are looking at is moving toward the right, the impact point of the wind would be left of the original appearing target.

You might be interested in

The platform height for Olympic divers is 10 m. A 60 kg diver steps off the platform to begin his dive.

azamat

Answer:

a) Ep = 5886[J]; b) v = 14[m/s]; c) W = 5886[J]; d) F = 1763.4[N]

Explanation:

a)

The potential energy can be found using the following expression, we will take the ground level as the reference point where the potential energy is equal to zero.

$E_{p} =m*g*h\\where:\\m = mass = 60[kg]\\g = gravity = 9.81[m/s^2]\\h = elevation = 10 [m]\\E_{p}=60*9.81*10\\E_{p}=5886[J]$

b)

Since energy is conserved, that is, potential energy is transformed into kinetic energy, the moment the harpsichord touches water, all potential energy is transformed into kinetic energy.

$E_{p} = E_{k} \\5886 =0.5*m*v^{2} \\v = \sqrt{\frac{5886}{0.5*60} }\\v = 14[m/s]$

c)

The work is equal to

W = 5886 [J]

d)

We need to use the following equation and find the deceleration of the diver at the moment when he stops his velocity is zero.

$v_{f} ^{2}= v_{o} ^{2}-2*a*d\\where:\\d = 2.5[m]\\v_{f}=0\\v_{o} =14[m/s]\\Therefore\\a = \frac{14^{2} }{2*2.5} \\a = 39.2[m/s^2]$

By performing a sum of forces equal to the product of mass by acceleration (newton's second law), we can find the force that acts to reduce the speed of the diver to zero.

m*g - F = m*a

F = m*a - m*g

F = (60*39.2) - (60*9.81)

F = 1763.4 [N]

3 0

2 years ago

What happens to the atoms that make up hydrogen fuel as it burns?

ale4655 [162]

Answer:

Get turned Into Water.

Explanation:

Combustion of Hydrogen involves combining oxygen and hydrogen essentially so when oxygen and hydrogen combine water is produced, following chemical equation describes this process.

$H_{2} +O_{2}=2(H_{2}O)$ .

Resulting product is two molecules of water.

4 0

3 years ago

A 95.0 kg satellite moves on a circular orbit around the Earth at the altitude h=1.20*103 km. Find: a) the gravitational force e

Natali5045456 [20]

Answer:

a) $F = 660.576\,N$ , b) $a_{c} = 6.953\,\frac{m}{s^{2}}$ , c) $v \approx 7255.423\,\frac{m}{s}$ , $\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$ , d) $T \approx 1.821\,h$

Explanation:

a) The gravitational force exerted by the Earth on the satellite is:

$F = G\cdot \frac{m\cdot M}{r^{2}}$

$F = \left(6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}} \right)\cdot \frac{(95\,kg)\cdot (5.972\times 10^{24}\,kg)}{(7.571\times 10^{6}\,m)^{2}}$

$F = 660.576\,N$

b) The centripetal acceleration of the satellite is:

$a_{c} = \frac{660.576\,N}{95\,kg}$

$a_{c} = 6.953\,\frac{m}{s^{2}}$

c) The speed of the satellite is:

$v = \sqrt{a_{c}\cdot R}$

$v = \sqrt{\left(6.953\,\frac{m}{s^{2}} \right)\cdot (7.571\times 10^{6}\,m)}$

$v \approx 7255.423\,\frac{m}{s}$

Likewise, the angular speed is:

$\omega = \frac{7255.423\,\frac{m}{s} }{7.571\times 10^{6}\,m}$

$\omega = 9.583\times 10^{-4}\,\frac{rad}{s}$

d) The period of the satellite's rotation around the Earth is:

$T = \frac{2\pi}{\left(9.583\times 10^{-4}\,\frac{rad}{s} \right)} \cdot \left(\frac{1\,hour}{3600\,s} \right)$

$T \approx 1.821\,h$

6 0

2 years ago

The binding energy of a nucleus is the energy equivalent of the mass discrepancy.

Dima020 [189]

ANSWER IS NOT FALSE IT IS True

6 0

2 years ago

If the velocity of an object is zero at one instant, what is true about the acceleration of that object

Charra [1.4K]

Answer:

I. The acceleration could be positive.

II. The acceleration could be zero.

III. The acceleration could be negative.

Explanation:

Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.

Mathematically, velocity is given by the equation;

$Velocity = \frac{distance}{time}$

If the velocity of an object is zero at one instant, then the acceleration of that object may have the following characteristics;

I. The acceleration could be positive.

II. The acceleration could be zero.

III. The acceleration could be negative.

In physics, acceleration can be defined as the rate of change of the velocity of an object with respect to time.

This simply means that, acceleration is given by the subtraction of initial velocity from the final velocity all over time.

Mathematically, acceleration is given by the equation;

$Acceleration (a) = \frac{final \; velocity - initial \; velocity}{time}$

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

Where,

a is acceleration measured in $ms^{-2}$

v and u is final and initial velocity respectively, measured in $ms^{-1}$

t is time measured in seconds.

8 0

2 years ago