You throw a ball vertically upward, and as it leaves your hand, its speed is 10.0 m/s.

1.what is gravity?(in your own words)

olga_2 [115]

Answer:

1.Gravity is an invisible force that pulls objects toward each other

2.The suns gravity pulls the planet towards the sun which changes the straight line of direction into a curve

That keeps the planet moving in an orbit around the sun

hope it helped:)

8 0

2 years ago

A container of gas molecules is at a pressure of 2 atm and has amass density of 1.7 grams per liter. All of the molecules in the

Tpy6a [65]

Answer:

The speed of nitrogen molecule is 1.87 m/s.

Explanation:

Given that,

Pressure = 2 atm

Density = 1.7 grams/liter

Atomic weight = 28 grams

We need to calculate the temperature

Using formula of idea gas

$PV=nRT$

$P=\dfrac{WRT}{VM}$

$P=\dfrac{\rho RT}{M}$

$T=\dfrac{PM}{\rho R}$

Put the value into the formula

$T=\dfrac{2\times28}{1.7\times0.0821}$

$T=401.2\ K$

We need to calculate the speed of nitrogen molecule

Using formula of RMS speed

$V_{rms}=\sqrt{\dfrac{3RT}{M}}$

$V_{rms}=\sqrt{\dfrac{3\times0.0821\times401.2}{28}}$

$V_{rms}=1.87\ m/s$

Hence, The speed of nitrogen molecule is 1.87 m/s.

5 0

3 years ago

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

5 0

3 years ago

A person is running at 2 m/s and 20 s later is running at 22 m/s. What is

AysviL [449]

I think it’s 1 m/s^2

8 0

3 years ago

A car traveling east increases its velocity from 4 m/s to 19 m/s in 5 s. What is the acceleration of the car?

Virty [35]

It’s going 3m/s. If we have 5 seconds to work with then we can find the acceleration by adding speed and how fast it going every second. So like the 7-10-13-16-19 so we go 3m/s faster ever second

3 0

3 years ago