Which are the three types of mutations?

Which of the following is equal to impulse? A. Ft B. m/s C. pt D. mv

tankabanditka [31]

Answer: A (Ft)

Explanation: The impulse experienced by the object equals the change in momentum of the object. In equation form, F • t = m • Δ v

5 0

3 years ago

is the light transmitted by a filter brighter than, less brighter than, or the same brightness as the light that hits it? explai

vichka [17]

The light transmitted by a filter is less bright than the light that hits it.

A filter only absorbs/removes part of the light that hits it. That means it removes some energy from the incident light. The remaining light carries less energy, so it must be less bright.

5 0

4 years ago

What force pulls the moon towards the earth?

yawa3891 [41]

Gravity pulls the moon towards the earth.

3 0

3 years ago

the 200 g baseball has a horizontal velocity of 30 m/s when it is struck by the bat, B, weighing 900 g, moving at 47 m/s. during

Ivanshal [37]

Solution :

Given :

Mass of the baseball, m = 200 g

Velocity of the baseball, u = -30 m/s

Mass of the baseball after struck by the bat, M = 900 g

Velocity of the baseball after struck by the bat, v = 47 m/s

According to the conservation of momentum,

$Mv+mu=Mv_1+mv_2$

(900 x 47) + (200 x -30) = (900 x $v_1$ ) + (200 x $v_2$ )

36300 = (900 x $v_1$ ) + (200 x $v_2$ )

$9v_1 + 2v_2 = 363$ ..............(i)

$9v_1 = 363 - 2v_2$

$v_1=\frac{363 - 2v_2}{9}$

The mathematical expression for the conservation of kinetic energy is

$\frac{1}{2}Mv^2+\frac{1}{2}mu^2 = \frac{1}{2}Mv_1^2+\frac{1}{2}mv_2^2$

$\frac{1}{2}(900)(47)^2+\frac{1}{2}(200)(-30)^2 = \frac{1}{2}(900)v_1^2+\frac{1}{2}(200)v_2^2$ ................(ii)

$(9)(14)^2+(2)(-30)^2 = (9)v_1^2+2v_2^2$

$21681 = 9v_1^2+2v_2^2$

Substituting (i) in (ii)

$21681= 9\left( \frac{363-2v_2}{9}\right)^2+2v_2^2$

$(363-2v_2)^2+18v_2^2=195129$

$(363)^2+18v_2^2-2(363)(2v_2)+(363)^2-195129=0$

$22v_2^2-145v_2-63360=0$

Solving the equation, we get

$v_2=96 \ m/s, -30 \ m/s$

The negative velocity is neglected.

Therefore, substituting 96 m/s for $v_2$ in (i), we get

$v_1=\frac{363-(2 \times 96)}{9}$

= 19

Thus, only impulse of importance is used to find final velocity.

8 0

3 years ago

In lab you find that a 1-kg rock suspended above water weighs 10 N. When the rock is suspended beneath the surface of the water,

pav-90 [236]

Answer:

a. 2N

b. 12N

c. 20N

Explanation:

Dtaa given,

Weight of rock in air=10N

weight of rock in water =8N

a. When the rock is suspend in water, it will displace some water from thr container since Buoyant force is the acts upward and equals in magnitude to the volume of water displays,

The buoyant force is calculated as 10N-8N=2N

b. The scale reading will be the sum of the buoyant force and the weight of the container

scale reading =10N+2N=12N

c. When the rock rest at the bottom of the container, the weight of the rock(10N) is acting downward and the weight of the container also is acting downward. Hence the sum of the two weight gives the reading on the scale

scale reading=10N+10N=20N

4 0

3 years ago