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3 years ago

Someone please help me with this physics!

Physics

1 answer:

Taya2010 [7]3 years ago

8 0

When it leaves the hand, it's rising at 18.2 m/s. After 1 second, gravity has slowed it to (18.2 - 9.8) = 8.4 m/s, still rising. Its average speed during the first second is 1/2 of (18.2 + 8.4) = 13.3 m/s. Rising for 1 second at an average speed of 13.3 m/s, it rises 13.3m from the hand.

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3 years ago

The length and width of a rectangular room are measured to be 3.92 ± 0.0035 m and 3.15 ± 0.0055 m. In this problem you can appro

Pavel [41]

Answer:

A) $A=12.2480\ m^2$

B) $12.2480\pm 0.1029\ m^2$

Explanation:

<u>Given:</u>

Length of the room $l= 3.92 ± 0.0035$

Width of the room $w= 3.15 ± 0.0055$

A) Let A be the area of the room

$A=l\times w\\A=3.92\times3.15\\A=12.2480\ \rm m^2$

B)We will calculate uncertainty in each dimension

%uncertainty in length $=\dfrac{0.0035}{3.92}\times 100=0.0892\ %$

%uncertainty in width = $\dfrac{0.0055}{3.15}\times 100=0.0174%$

The uncertainty in area will be sum of uncertainty in length and width

%uncertainty in Area= %uncertainty in length + %uncertainty in width

%uncertainty in Area $=0.0892\ % + 0.0174\ %$

%uncertainty in Area=0.0106

Uncertainty in Area $=0.0106\times 12.2480=0.1029\ \rm m^2$

There Area is $12.2480 ± 0.1029\ \rm m^2$

7 0

3 years ago

If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite d

Yuliya22 [10]

Incomplete question as the mass of baseball is missing.I have assume 0.2kg mass of baseball.So complete question is:

A baseball has mass 0.2 kg.If the velocity of a pitched ball has a magnitude of 44.5 m/sm/s and the batted ball's velocity is 55.5 m/sm/s in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat.

Answer:

ΔP=20 kg.m/s

Explanation:

Given data

Mass m=0.2 kg

Initial speed Vi=-44.5m/s

Final speed Vf=55.5 m/s

Required

Change in momentum ΔP

Solution

First we take the batted balls velocity as the final velocity and its direction is the positive direction and we take the pitched balls velocity as the initial velocity and so its direction will be negative direction.So we have:

$v_{i}=-44.5m/s\\v_{f}=55.5m/s$