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3 years ago

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Which are complete descriptions of the momentum of an object? Circle all that apply. a. 2.0 kg/s b. 7.2 kg•m/s, right c. 6.1 kg•

m/s2, down d. 4.2 m/s, east e. 1.9 kg•m/s, west f. 2.3 kg•m/s

1 answer:

Paul [167]3 years ago

8 0

<span>mutations due to exposure to chemicals</span>

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PLS THIS IS DUE IN 2 MINUTES

luda_lava [24]

Answer:

The toy car

Explanation:

the real car is parked so yeah but maybe in some way technically the real car has more "momentum"

7 0

2 years ago

How does the total amount of energy change as the spring goes up and down?

NeX [460]

The total amount of energy always stays the same because of the law of conservation of energy, meaning that there is always the same amount of energy, it just gets turned into different forms like potential energy, kinetic energy, sound, thermal, etc.

7 0

2 years ago

A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth

Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

$\tan\theta=\dfrac{8}{3}$

$\thta=\tan^{-1}\dfrac{8}{3}$

$\theta=69.44^{\circ}$

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

$mg=2T\cos\theta$

Put the value into the formula

$3\times9.8=2T\times\cos69.44$

$T=\dfrac{3\times9.8}{2\times\cos69.44}$

$T=41.85\ N$

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0

3 years ago

Read 2 more answers

A car is running at the speed of 45km/hr see a child 25 meter ahead and suddenly apllies a brakes. If the retradation of the car

kenny6666 [7]

Answer:

The car stops in 7.78s and does not spare the child.

Explanation:

In order to know if the car stops before the distance to the child, you take into account the following equation:

$x=x_o+v_ot-\frac{1}{2}at^2$ (1)

vo: initial speed of the car = 45km/h

a: deceleration of the car = 2 m/s^2

t: time

xo: initial distance to the child = 25m

x: final distance to the child = 0m

It is necessary that the solution of the equation (1) for time t are real.

You first convert the initial speed to m/s, then replace the values of the parameters and solve the quadratic polynomial for t:

$45\frac{km}{h}*\frac{1h}{3600s}*\frac{1000m}{1km}=12.5\frac{m}{s}$

$0=25+12.5t-2t^2\\\\2t^2-12.5t-25=0\\\\t_{1,2}=\frac{-(-12.5)\pm \sqrt{(-12.5)^2-4(2)(-25)}}{2(2)}\\\\t_{1,2}=\frac{12.25\pm 18.87}{4}\\\\t_1=7.78s\\\\t_2=-1.65s$

You take the first value t1 because it has physical meaning.

The solution for t is real, then, the car stops in 7.78s and does not spare the child.

4 0

3 years ago

A block of mass m is pushed up against a spring, compressing it a distance x, and the block is then released. The spring project

Gwar [14]

Answer:

$x' = 10 x$

Explanation:

By energy conservation we know that spring energy is converted into kinetic energy of the block

so we will have

$\frac{1}{2}kx^2 = \frac{1}{2}mv^2$

so we will have

$v = \sqrt{\frac{k}{m}}(x)$

now we will have same thing for another mass 4m which moves out with speed 5v

so we have

$5v = \sqrt{\frac{k}{4m}}(x')$

now from above two equations we have

$\frac{5v}{v} = \frac{x'}{2x}$

so we have

$x' = 10 x$

3 0

2 years ago