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3 years ago

I have two questions I need help with.

1 answer:

4 0

1). The Earth would need much much much much much more mass than it has in order to see any change in our orbital period around the sun. An increase of 20% won't jhave any effect.

3). This question isn't clear. "years" isn't a unit of length or distance. It can't describe a change in our distance from the sun, so we don't know what this question is actually asking.

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I NEED HELP SOLVING THIS!!!!!!!!!!!!

Neko [114]

Answer:

Yes

Explanation:

The given parameters are;

The speed with which the fastball is hit, u = 49.1 m/s (109.9 mph)

The angle in which the fastball is hit, θ = 22°

The distance of the field = 96 m (315 ft)

The range of the projectile motion of the fastball is given by the following formula

$Range = \dfrac{u^2 \times sin(2\cdot \theta)}{g}$

Where;

g = The acceleration due to gravity = 9.81 m/s², we have;

$Range = \dfrac{49.1^2 \times sin(2\times22^{\circ})}{9.81} \approx 170.71 \ m$

Yes, given that the ball's range is larger than the extent of the field, the batter is able to safely reach home.

7 0

3 years ago

If you fired a rifle straight upwards at 1000 m/s, how far up will the bullet get?

nadya68 [22]

Answer:

h = 51020.40 meters

Explanation:

Speed of the rifle, v = 1000 m/s

Let h is the height gained by the bullet. It can be calculated using the conservation of energy as :

$\dfrac{1}{2}mv^2=mgh$

$h=\dfrac{v^2}{2g}$

$h=\dfrac{(1000\ m/s)^2}{2\times 9.8\ m/s^2}$

h = 51020.40 meters

So, the bullet will get up to a height of 51020.40 meters. Hence, this is the required solution.

4 0

3 years ago

1) Describe how the atoms of a solid differ from the atoms of a liquid.

lisabon 2012 [21]

Answer:

liquid are close together with no regular arrangement. solid are tightly packed, usually in a regular pattern.

6 0

4 years ago

2200 kg semi truck driving down the highway has lost control. The truck rolls across the median and into oncoming traffic. The t

serious [3.7K]

Answer:

The semi truck travels at an initial speed of 69.545 meters per second downwards.

Explanation:

In this exercise we see a case of an entirely inellastic collision between the semi truck and the car, which can be described by the following equation derived from Principle of Linear Momentum Conservation: (We assume that velocity oriented northwards is positive)

$m_{S}\cdot v_{S}+m_{C}\cdot v_{C} = (m_{S}+m_{C})\cdot v$ (1)

Where:

$m_{S}$ , $m_{C}$ - Masses of the semi truck and the car, measured in kilograms.

$v_{S}$ , $v_{C}$ - Initial velocities of the semi truck and the car, measured in meters per second.

$v$ - Final speed of the system after collision, measured in meters per second.

If we know that $m_{S} = 2200\,kg$ , $m_{C} = 2000\,kg$ , $v_{C} = 45\,\frac{m}{s}$ and $v = -15\,\frac{m}{s}$ , then the initial velocity of the semi truck is: