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3 years ago

PLEASE HELP ME PLEASE! BRAINLEST

Physics

1 answer:

Vadim26 [7]3 years ago

6 0

Answer:

10kg

Explanation:

Weight is "how much does gravity drag this down".

Mass is "how much matter is there here".

The relation is:

$F_g = mg$

where $F_g$ is the weight, $m$ is the mass and $g$ is the gravitational acceleration (roughly equal to 10N/kg on Earth).

From the task we know that:

$F_g = 100N\\g = 10\frac{N}{kg}$

So let's input it into the relation:

$100N = m\cdot 10\frac{N}{kg}\\10N = m \cdot 1\frac{N}{kg}\\10N \cdot \frac{kg}{N} = m\\~\\m = 10kg$

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Difference between physics and science

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Physics is branch of science which deals with the properties of matter and their energies. And coming to science it is a systemic enterprise that builds and organizes knowledge in t eatable explanations and predictions about the universe.

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3 years ago

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A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car

viktelen [127]

This question is not complete, the complete question is;

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance.

In the following cases, the car initially has a total mass (car and contents) of 170 kg and is traveling east with a velocity of magnitude 4.60 m/s.

Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks.

Part A

An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity.

Part B

An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.

Answer:

Part A) the final velocity of the car is 4.6 m/s

Part B) the final velocity of the car is 5.28 m/s

Explanation:

Given the data in the question;

Total mass (m₁+m₂) = 170 kg

velocity of magnitude Vx = 4.60 m/s

PART A)

An object with a mass of 22.0 kg is thrown sideways out of the car with a speed of 2.50 m/s relative to the car's initial velocity,

i.e

m₂ = 22.0 kg

so m₁ = 170 - 22 = 148 kg

so, we apply conservation of momentum

since the object thrown out of the car, it has nothing to do with the car's velocity.

(m₁+m₂)Vx = m₁Vx₁ + m₂Vx₂

we substitute

(170)4.60 = 148Vx₁ + 22(4.60)

782 = 148Vx₁ + 101.2

148Vx₁ = 782 - 101.2

148Vx₁ = 680.8

Vx₁ = 680.8 / 148

Vx₁ = 4.6 m/s

Therefore, the final velocity of the car is 4.6 m/s

Part B)

An object with a mass of 22.0 kg is thrown backward out of the car with a velocity of 4.60 m/s relative to the initial motion of the car.

Vx = V(m₁+m₂) / ((m₁+m₂) - m₁)

we substitute

Vx = 4.60(170) / ((170) - 22)

Vx = 782 / 148

Vx = 5.28 m/s

Therefore, the final velocity of the car is 5.28 m/s

7 0

3 years ago

A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 3.00 m/s, and

vladimir1956 [14]

Answer:

a) t= 0.92 s

b) h = 0.46 m

c) v = -6.04 m/s

Explanation:

In order to find the total time that the feet are in the air, we must add two times:
1) time needed to reach to the maximum height (t₁)
2) time from when starts to fall from the maximum height until her feet hit the water (t₂)
In order to get t₁, we need to take into account that at her highest point, the vertical speed will be zero.
Taking for granted the value for the acceleration due to gravity,

g = -9.8 m/s2, we can apply the definition of acceleration, and replacing by the givens, we can find t₁ as follows:

$t_{1} = \frac{v_{o}}{g} = \frac{3.0m/s}{9.8m/s2} = 0.3 s (1)$

In order to find t₂, we need to find first the highest point above the board, which is indeed what is asked for in b).
We can use the following kinematic equation, taking into account that at the highest point, the final velocity vf will be zero.
The equation can be written as follows:

$v_{f} ^{2} - v_{o} ^{2} = 2*g*\Delta h (2)$

Replacing by the givens, and solving for Δh, we get:

$\Delta h = \frac{v_{o}^{2}}{2*g} = \frac{(3.0m/s)^{2} }{2*9.8m/s2} = 0.46 m (3)$

The total height over the water will be just the sum of the takeoff point (1.40 m over the water) and the value we found in (3):
H = 1.40 m + 0.46 m = 1.86 m
Now, we can use the equation that relates the vertical displacement with the time, remembering that v₀=0, as follows:

$H = \frac{1}{2}*g*t^{2} (4)$

Replacing by the givens and the value found for H in (4), and solving for t, we get the value of t₂, as follows:

$t_{2} = \sqrt{\frac{2*H}{g} } = \sqrt{\frac{2*1.86m}{9.8m/s2} } = 0.62 s (5)$

The total time will the sum of t₁ and t₂:
t = 0.3 s + 0.62 s = 0.92 s (6)

As we have just found, the highest point above the board was at 0.46m above the takeoff point, so the highest point above the board is just 0.46 m.

In order to find the velocity when her feet hit the water, we can use the same equation (2), taking into account that v₀=0, and Δh = H= -1.86m.
Solving (2) for vf, we get:

$v_{f} = - \sqrt{2*g*H} = -\sqrt{2*9.8m/s2*1.86m} = -6.04 m/s (7)$

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3 years ago

What force is represented by the vector? friction gravity normal force push

Ket [755]

Answer:

gravity

Explanation:

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Answer:

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PLEASE HELP ME PLEASE! BRAINLEST ​

PLEASE HELP ME PLEASE! BRAINLEST