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Eva8 [605]
3 years ago
9

An object is realsed From rest . how far does it fall during the second second of fall?​

Physics
1 answer:
Savatey [412]3 years ago
6 0

Answer:

14.715 m

Explanation:

Assume that the acceleration due to gravity is 9.81 m/s^2 downwards, take downwards as positive

First second:

v = u + at

v = 9.81 m/s

Second second:

s = ut + (1/2)at^2

s = 9.81(1) + (1/2)(9.81)(1)^2

s = 14.715 m

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Acceleration is a change in _____.
Rus_ich [418]

Answer:

direction

Explanation:

4 0
3 years ago
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If an object weighs 300 N on earth, what is it’s mass on the moon?
inna [77]

Answer:

The mass of the object on the Moon (and anywhere else) is about 30.61kg. Please see more detail below.

Explanation:

Weight is the gravitational force exerted on the object and is a function of mass and gravitational acceleration:

(weight) = (mass) x (gravitational acceleration)

We are to find the mass, knowing the weight on Earth to be 300N:

(mass) = (weight on Earth) / (gravitational acceleration on Earth) = 300N / 9.8 m/s^2 = 30.61 kg

The mass of the object is 30.61kg.

The mass of the object is independent of gravity. Therefore the answer to the question "What is its mass on the Moon" is 30.61kg.

If the question were what is its weight on the Moon, the answer would be

(weight on Moon) = (mass) x (grav.accel. on Moon) = 30.61kg x 1.62 m/s^2 = 49.59N

which is about 1/6 of the object's weight on the Earth.

4 0
3 years ago
A 20.94 g sample of an unknown metal is heated to 99.4 oC in a hot water bath. The metal sample is transferred to a calorimeter
MAXImum [283]

Answer:

1092 J

Explanation:

From the question,

Assuming no heat is lost to the surrounding

Heat lost by the metal = Heat gained by water.

H' = Cm(t₂-t₁)................. Equation 1

Where H' = Heat lost by the metal, C = specific heat capacity of water,  m = mass of water, t₂ = Final temperature of the mixture, t₁ = Initial temperature of water

But,

m = D'v................... Equation 2

Where D' = Density of water, v = volume of water.

Given: D' = 1000 kg/m³, v = 100 mL = 100/1000000 = 0.0001 m²

Substitute into equation 2

m = 1000(0.0001)

m = 0.1 kg.

Also given: C = 4200 J/kg.°C, t₁ = 22 °C, 24.6 °C

Substitute into equation 1

H' = 4200×0.1×(24.6-22)

H' = 420(2.6)

H' = 1092 J.

Hence the heat (q) lost by the metal = 1092 J

3 0
4 years ago
1. The center of gravity of both cats and humans is roughly in the thoracic region. However, they each bear their weight very di
Mama L [17]

Answer:

Major, weight-bearing structures are the bones of the body that are strong and dense to be able to bear the weight of the body. The major, weight-bearing structures of cat and human skeletons are :

Human skeleton: The body weight of an individual is on his pelvic girdles that are attached to the bones of lower limbs. Thigh bones, leg bones, and bones of feet comprise lower limbs The lower limbs consist of the thigh, the leg, and the foot.

Cat skeleton:  cats are quadrupedal so it bears all the body weight on shoulders and legs that includes the Scapula and pelvis.

6 0
3 years ago
What two processes practiced by scientists increase the likelihood of a successful outcome in science?
Amanda [17]

Answer:

ULTIMATE CORRECT ANSWER

collaboration and communication

Explanation:

5 0
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