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3 years ago

Whenever an object is moving at a constant rate, the value(s) that equal zero is (are):

Physics

2 answers:

dolphi86 [110]3 years ago

5 0

Answer:

d) all of the above

Explanation:

the speed, velocity, and acceleration all changes to 0 due to no motion.

Xelga [282]3 years ago

3 0

Whenever an object is moving at a constant rate . . .

a). its speed is that constant rate.

b). Its <em>acceleration MAY BE zero</em>, if it's also moving in a straight line.

c). Its velocity is that constant rate if it's moving in a straight line. Otherwise, its velocity could have many different values, depending on the path it's following.

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What is the significance of Nucleotides in Chromosomes?

fgiga [73]

Answer:

it comprises of the DNA/RNA bipolymer molecules

3 0

3 years ago

Read 2 more answers

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$

Thus, the final velocity of thrower is $3.88~m/s$ and that for the catcher is $0.029~m/s$ .

8 0

3 years ago

How much net force is required to accelerate a 2000 kg car at 3.00 m/s^2

andrezito [222]

The net force required to accelerate a car is 6000 N.

Force is defined as the product of the mass and acceleration of the body. Force is used to changing the velocity that is to accelerate an object or a body of a particular mass. The unit of Force is Newton or kg m/s^2.

The formula used to calculate the net force is :

F = ma

where, F = Force

m = mass = 2000 kg

a = acceleration = 3.00 m/s^2

∴ F = 2000*3

F = 6000 N

Thus, to accelerate the car at 3.00 m/s^2 of mass 2000 kg net force required is 6000 N.

To learn more about force,

brainly.com/question/1046166

6 0

1 year ago

If a bullet travels at 593.0 m/s, what is its speed in miles per hour?

Ksenya-84 [330]

We have $1 \; mile = 1609.34\; meters$ . So, $1 \; meter = \frac{1}{1609.34} \;mile$ .

$1 \; hour = 3600 \; s$ . So $1 \; s = \frac{1}{3600} \; hour$ .

Thus we can convert the units of the given quantity.

That is,

$593\;m/s=593\;\frac{1/1609.34}{1/3600} \;miles/hour\\
593\;m/s=1,326.51\;miles/hour$ .

The quantity is converted to the required units.

7 0

3 years ago

An object with a mass m slides down a rough 370 inclined plane where the coefficient of kinetic friction is 0.20. If the plane i

Svetllana [295]

Answer:

$v \approx 9.312\,\frac{m}{s}$

Explanation:

The Free Body Diagram of the system is presented in the image attached below. The final speed is determined by means of the Principle of Energy Conservation and the Work-Energy Theorem:

$K_{A} + U_{g,A} = K_{B} + U_{g,B} + W_{loss}$