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2 years ago

The diagram shows two balls before they collide.

Physics

2 answers:

mihalych1998 [28]2 years ago

6 0

Answer:

B. 0.2 kg x m/s

Explanation:

Ronch [10]2 years ago

4 0

Answer:

The Answer is B)0.2 kg • m/s

Explanation:

I made a 100 on my test. Sorry if I'm late but hope I helped.

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Depending on circumstances, there are times when divers may wish to be positively, negatively or neutrally buoyant. True. False.

ser-zykov [4K]

Answer:

True

Explanation:

Buoyancy is the most important factors for divers. All they do underwater is to observe the life down there but they also have some other work. However, divers may want to be negatively buoyant when they want to go on deep exploration. When they reach a destination, they may want to observe and neutral buoyancy then will be useful. When they want to go back on surface, they’ll utilize positive buoyancy.

7 0

3 years ago

How can you find the mass of a tissue box?

a_sh-v [17]

Explanation:

To find the volume of a box of Tissue (also known as a rectangular prism) just multiply the area of the base (length times width) by the height. The formula is V = l• w• h.

7 0

2 years ago

In a ballistics test, a 24 g bullet traveling horizontally at 1200 m/s goes through a 31-cm-thick 320 kg stationary target and e

Zanzabum

Answer:

The velocity is $v_t = 0.02175 \ m/s$

Explanation:

From the question we are told that

The mass of the bullet is $m_b = 0.024 \ kg$

The initial speed of the bullet is $u_b = 1200 \ m/s$

The mass of the target is $m_t = 320 \ kg$

The initial velocity of target is $u_t = 0 \ m/s$

The final velocity of the bullet is is $v_b = 910 \ m/s$

Generally according to the law of momentum conservation we have that

$m_b * u_b + m_t * u_t = m_b * v_b + m_t * v_t$

=> $0.024 * 1200 + 320 * 0 = 0.024 * 910 + 320 * v_t$

=> $v_t = 0.02175 \ m/s$

3 0

3 years ago

A dockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-N horiz

Nataliya [291]

1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

$F_s = \mu_s mg$

where

$\mu_s$ is the static coefficient of friction

m = 21 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

$F_s=73 N$

Using this information into the previous equation, we can find the coefficient of static friction:

$\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355$

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

$F_k = \mu_k mg$ (1)

where

$\mu_k$ is the coefficient of kinetic friction

There is a horizontal force of

F = 55 N

pushing the crate. Moreover, the speed of the crate is constant: this means that the acceleration is zero, a = 0.

So we can write Newton's second law as

$F-F_k = ma = 0$

And by substituting (1), we can find the value of the coefficient of kinetic friction:

$F-\mu_k mg = 0\\\mu_k = \frac{F}{mg}=\frac{55 N}{(21 kg)(9.8 m/s^2)}=0.267$

5 0

3 years ago

WILL MARK BRAINLIEST! The driver of a pickup truck going 120 km/h applies the brakes, giving the truck a uniform deceleration of

Rudik [331]

First put the speed in m/s. 120km/h = 33.33m/s. Now the position function is the integral of velocity, and velocity is in turn the integral of acceleration. The velocity is:
$v= \int\limits^{} _ {} {-5} \, dt =-5t+33$
Now we integrate this expression to get the position. The constant of integration will be the distance the truck travels.
$s= \int\limits^{} {-5t+33} \, dx =- \frac{5}{2}t^2+33.33t-35=0$
Here we set the distance, 35m as negative because I assumed the stopping point of the truck is the origin. Putting t=0 shows it starts at -35m.
Now solve the following equation for the time, t using the quadratic equation:
$s=- \frac{5}{2}t^2+33.33t-35=0$
and choose the value t=1.149s

7 0

3 years ago