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monitta
3 years ago
9

An object with an initial velocity of 10 m/s accelerates at a rate of 3.5 m/s2 for 8 seconds. How far will it have traveled duri

ng that time?
Physics
1 answer:
ICE Princess25 [194]3 years ago
4 0

Answer:

Distance, d = 192 meters

Explanation:

We have,

Initial velocity of an object is 10 m/s

Acceleration of the object is 3.5 m/s²

Time, t = 8 s

We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :

d=ut+\dfrac{1}{2}at^2

d=10\times 8+\dfrac{1}{2}\times 3.5\times 8^2\\\\d=192\ m

So, the distance travelled by the object is 192 meters.

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a hot iron bar is placed 100ml 22C water. the water temperature rises to 32C. how much heat did the water gain and how much heat
Sergeu [11.5K]

Answer:

  • Water gained: 10
  • Iron lost: -10

Explanation:

Given: Hot iron bar is placed 100ml 22C water, the water temperature rises to 32C

To find: How much heat the water gain, how much heat did the iron bar lost

Formula:Q = change T x C x M

Solve:

<u>How much heat water gained</u>

Initial heat = 22, then rose to 32. To find how much heat the water gained, simply subtract the current heat by the initial heat.

                                              32 - 22 = 10

The water gained 10 amounts of heat.

<u>How much heat Iron lost</u>

Current heat = 32, then dropped to 22. To find how much heat the Iron lost, simply subtract the initial heat by the current heat.

                                                   22 - 32 = -10

The Iron lost -10 amounts of water.      

4 0
3 years ago
The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth e
djverab [1.8K]
The gravitational field is the Force divided by the mass

Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.

g = F / m


g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg

Note that the unit N/kg is equivalent to m/s^2

 
8 0
4 years ago
An orbiting satellite can become charged by the photoelectric effect when sunlight ejects electrons from its outer surface. Sate
Rufina [12.5K]

Answer:

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

Explanation:

Given data

Φ = 5.32 eV

to find out

the longest wavelength

solution

we know that

hf = k(maximum) +Ф   ...............1

here we consider k(maximum ) will be zero because photon wavelength max when low photon energy

so hf = 0

and hc/ λ = +Ф

so λ = hc/Ф  ................2

now put value hc = 1240 ev nm and Φ = 5.32 eV

so hc = 1240 / 5.32

hc = 233 nm

the longest wavelength of incident sunlight that can eject an electron from the platinum is 233 nm

8 0
3 years ago
A 0.400-kg object is swung in a circular path and in a vertical plane on a 0.500-m-length string. If the angular speed at the bo
Talja [164]

Answer:

T = 16.72 N

Explanation:

When the object is swung in a circular path, and in a vertical plane, there are two forces external to the object acting on it at any time: the gravity (which is always downward) and the tension in the string (which always points towards the center of the circle).

At the bottom of the circle, the tension is directly upward, so these two forces, are opposite each other, and the difference between them is the centripetal force , which at this point, keeps the object swinging in a circle.

This is the point of the trajectory where T is maximum.

We can apply Newton's 2nd Law, choosing an axis vertical (y-axis) being the upward direction the positive one, as follows:

T- m*g = m*a

The acceleration, at the bottom of the circle, is only normal (as there are no forces in the horizontal direction) , and is equal to the centripetal acceleration, as follows:

ac =  v² / r = ω²*r⇒ T- m*g = m*ω²*r

Replacing by the givens, we can solve for T as follows:

T = m* (ω²*r+g) = 0.4 kg*((8.00)² rad/sec²*0.5m)+9.8 m/s²) = 16.72 N

5 0
3 years ago
What is the relationship between the temperature of a star and its luminosity?
Sladkaya [172]
The Luminosity of a star is proportional to its Effective Temperature to the 4th power and its Radius squared.
3 0
3 years ago
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