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3 years ago

Explain how your weight would change if you were on a space ship traveling to the moon

1 answer:

4 0

The farther from earth's surface you are, the less gravitational attraction there is. So you will weigh less on a space ship because you have escaped the gravitational pull of earth. When you finally reach the moon, your weight is roughly one sixth what it is on earth.

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Which biome has the most variable year round temperature

skad [1K]

That should be the Grasslands ?

5 0

2 years ago

NEED HELP ASAP

Dafna11 [192]

Answers:

a) -2.54 m/s

b) -2351.25 J

Explanation:

This problem can be solved by the <u>Conservation of Momentum principle</u>, which establishes that the initial momentum $p_{o}$ must be equal to the final momentum $p_{f}$ :

$p_{o}=p_{f}$ (1)

Where:

$p_{o}=m_{1} V_{o} + m_{2} U_{o}$ (2)

$p_{f}=(m_{1} + m_{2}) V_{f}$ (3)

$m_{1}=110 kg$ is the mass of the first football player

$V{o}=-7 m/s$ is the velocity of the first football player (to the south)

$m_{2}=75 kg$ is the mass of the second football player

$U_{o}=4 m/s$ is the velocity of the second football player (to the north)

$V_{f}$ is the final velocity of both football players

With this in mind, let's begin with the answers:

a) Velocity of the players just after the tackle

Substituting (2) and (3) in (1):

$m_{1} V_{o} + m_{2} U_{o}=(m_{1} + m_{2}) V_{f}$ (4)

Isolating $V_{f}$ :

$V_{f}=\frac{m_{1} V_{o} + m_{2} U_{o}}{m_{1} + m_{2}}$ (5)

$V_{f}=\frac{(110 kg)(-7 m/s) + (75 kg) (4 m/s)}{110 kg + 75 kg}$ (6)

$V_{f}=-2.54 m/s$ (7) The negative sign indicates the direction of the final velocity, to the south

b) Decrease in kinetic energy of the 110kg player

The change in Kinetic energy $\Delta K$ is defined as:

$\Delta K=\frac{1}{2} m_{1}V_{f}^{2} - \frac{1}{2} m_{1}V_{o}^{2}$ (8)

Simplifying:

$\Delta K=\frac{1}{2} m_{1}(V_{f}^{2} - V_{o}^{2})$ (9)

$\Delta K=\frac{1}{2} 110 kg((-2.5 m/s)^{2} - (-7 m/s)^{2})$ (10)

Finally:

$\Delta K=-2351.25 J$ (10) Where the minus sign indicates the player's kinetic energy has decreased due to the perfectly inelastic collision

6 0

2 years ago

A race car traveling at 10 meters per second accelerates at 1.5 meters per second squared while moving a distance of 600 meters.

Mekhanik [1.2K]

Answer:

Explanation:

Givens

vi = 10 m/s

a = 1.5 m/s^2

d = 600 m

vf = ?

Formula

vf^2 = vi^2 + 2*a*d

Solution

vf^2 = 10^2 + 2*1.5 * 600

vf^2 = 100 + 1800

vf^2 = 1900

sqrt(vf^2) = sqrt(1900)

vf = 43.59 m/s

7 0

2 years ago

A projectile is fired vertically upwards and reaches a height of 78.4 m. Find the velocity of projection and the time it takes t

Musya8 [376]

Answer:

1.) U = 39.2 m/s

2.) t = 4s

Explanation: Given that the

height H = 78.4m

The projectile is fired vertically upwards under the acceleration due to gravity g = 9.8 m/s^2

Let's assume that the maximum height = 78.4m. And at maximum height, final velocity V = 0

Velocity of projections can be achieved by using the formula

V^2 = U^2 - 2gH

g will be negative as the object is moving against the gravity

0 = U^2 - 2 × 9.8 × 78.4

U^2 = 1536.64

U = sqrt( 1536.64 )

U = 39.2 m/s

The time it takes to reach its highest point can be calculated by using the formula;

V = U - gt

Where V = 0

Substitute U and t into the formula

0 = 39.2 - 9.8 × t

9.8t = 39.2

t = 39.2/9.8

t = 4 seconds.

7 0

3 years ago

How can two strengths be added together

Ket [755]

By using mind and dna

6 0

2 years ago