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3 years ago

This is for physical science, need help :/

Physics

1 answer:

marusya05 [52]3 years ago

6 0

1). From the frame of reference of a passenger on the airplane looking out of his window, the tree appears to be moving, at roughly 300 miles per hour toward the left of the picture.

2). The SI unit best suited to measuring the height of a building is the meter.

3). 'Displacement' is the straight-line distance and direction from the start-point to the end-point, regardless of the path that was followed to get there.

The ball started out in the child's hand, and it ended up 2 meters away from her in the direction of the wall. So the displacement of the ball from the beginning to the end of the story is: 2 meters toward the wall.

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Heat energy refers to the kinetic energy of molecules. heat can move in a number of different ways: when warm air rises causing

Anna71 [15]

Convection is the process that is said to occur

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3 years ago

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A motorist wishes to travel 40 kilometers at an average speed of 40 km/h. During the first 20 kilometers, an average speed of 40

allochka39001 [22]

poste en français s’il vous plaît

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3 years ago

A m = 94.2 kg object is released from rest at a distance h = 1.15134 R above the Earth’s surface. The acceleration of gravity is

OleMash [197]

Answer:

v= 4055.08m/s

Explanation:

This is a problem that must be addressed through the laws of classical mechanics that concern Potential Gravitational Energy.

We know for definition that,

$U = \frac{GMm}{r}$

We must find the highest point and the lowest point to identify the change in energy, so

Point a)

The problem tells us that an object is dropped at a distance of h = 1.15134R over the earth.

That is to say that the energy of that object is equal to,

$U_1=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(1.15134)(6.38*10^6)}$

$U_2= - 5.1180*10^9J$

Point B )

We now use the average radius distance from the earth.

$U_2=-\frac{(6.6738 * 10^{-11})(5.98 * 10^{24})(94.2)}{(6.38*10^6)}$

$U_2= -5.8925*10^9J$

Then,

$\Delta U = U_2 - U_1 = -5.1180*10^9J - ( -5.8925*10^9J)$

$\Delta U = 774.5*10^6$

By the law of conservation of energy we know that,

$\Delta U = \frac{1}{2}mv^2$

clearing v,

$v= \sqrt{2 \Delta U/m}$

$v= \sqrt{2*774.5*10^6 /94.2}$

$v= 4055.08m/s$

Therefore the speed of the object when it strikes the Earth’s surface is 4055.08m/s

8 0

3 years ago

Two in-phase loudspeakers that emit sound with the same frequency are placed along a wall and are separated by a distance of 5.0

IceJOKER [234]

Answer:

Explanation:

This is a case of interference of sound , akin to YDSE in optics .

Here, like interference dark and bright fringes, region of silence and intense sound will be formed due to destructive and constructive interference respectively.

Here d = distance between two sources = 5 m

D = distance of source and screen = 12m

position of first destructive interference

= λ D /2d

1 = λ 12 /2x 5

λ = 5 / 6 m

frequency = v / λ

= 343 x 6/ 5

= 411.6 Hz

7 0

3 years ago

By what factor would you need to adjust the length of a pendulum to make the period of 1/2 of what it used to be?

Irina-Kira [14]

Answer:

1/4

Explanation:

The period of a simple pendulum is:

T = 2π √(L/g)

At half the period:

T/2

= π √(L/g)

= 2π √(L/(4g))

= 2π √((L/4)/g)

So the length would have to be shortened by a factor of 1/4.

3 0

3 years ago