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2 years ago

Please Help! A ball is thrown straight up from the ground. What way does its acceleration point at the top?

Physics

1 answer:

QveST [7]2 years ago

3 0

Answer: A. Vertical

Explanation:

If the ball is thrown straight up from the ground (asuming the ground as height 0), this means its initial velocity is greater than zero.

While the ball rises and gains height, its velocity decreases until it reaches its maximum height where it stops (velocity equal to zero) and then it begins to fall to the ground.

Now, <u>during all the movement, the ball has an acceleration</u>, which is the acceleration due gravity (9.8 m/s^{2} on Earth), even at the top or maximum height (when the ball stops just for fraction of time), and this acceleration points vertical and downward to the Earth's center.

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A state patrol officer saw a car start from rest at a highway on-ramp. She radioed ahead to another officer 20 mi along the hig

Vikentia [17]

Answer:

We know from the basic speed distance relation that

$Speed=\frac{Distance}{Time}$

Since the car started from rest and it covered the distance between the 2 officer's in 19 minutes we have speed of the car

$Speed=\frac{Distance}{Time}\\\\Speed=\frac{20}{\frac{19}{60}}=63.16mph$

Which clearly exceeds the limit of $60\frac{mi}{hr}$

5 0

3 years ago

An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an

Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of $\gamma$

Using formula of relativistic energy

$E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

Put the value into the formula

$1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}$

Here, $\gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)$

$\gamma-1=0.01953$

$\gamma=0.01953+1$

$\gamma=1.01953$

We need to calculate the length

Using formula of length

$L'=\dfrac{L}{\gamma}$

Put the value into the formula

$L'=\dfrac{4}{1.01953}$

$L'=3.92\ m$

Hence, The length of the tube is 3.92 m.

8 0

3 years ago

What medical technique uses sound wave pulses and their reflections to map structures and organs inside the body?

garri49 [273]

The correct answer is “C” ultrasound. Hope this helps!

7 0

3 years ago

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As more lamps are put into a series circuit, the overall current in the circuit a. Increasesb. Decreasesc. Remains the same

erik [133]

Answer:

b. Decreases

Explanation:

The total resistance of a series circuit is equal to the sum of the individual resistances:

$R_T=R_1+R_2+...+R_n$ (1)

Therefore, as we add more lamps, the total resistance increases (because we add more positive tems in the sum in eq.(1).

The current in a circuit is given by Ohm's law:

$I=\frac{V}{R_T}$

where V is the voltage provided by the power source and $R_T$ is the total resistance. We notice that the current, I, is inversely proportional to the total resistance: therefore, when more lamps are added to the series circuit, the total resistance increases, and therefore the current in the circuit decreases.

8 0

3 years ago

If the current throught a 6 Ω resistor is 7 A, what is the power absorbed by the resistor? Answer to the nearest 1 W.

Stells [14]

Answer:

294 W

Explanation:

Electrical power: This can be defined as the rate at which electrical energy is used or dissipated. The S.I unit of electrical power is Watt(W).

The expression of electrical power is given as,

P = I²R................... Equation 1

Where P = power, I = current, R = Resistance.

Given: I = 7 A, R = 6 Ω

Substitute into equation 1

P = (7)²(6)

P = 294 W.

Hence the power power absorbed by the resistor is 294 W

3 0

3 years ago

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