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Katen [24]
3 years ago
14

PLEASE SOMEONE HELP ME PLEASE

Physics
1 answer:
serious [3.7K]3 years ago
4 0
Pe = m • g • h
= 100 • 9.8 • 20
= 19,600 or 1.96 x 10^4
You might be interested in
A SMART car can accelerate from rest to a speed of 28 m/s in 20s. What distance does it travel in this time?
sashaice [31]
Speed= distance/ time so distance = speed * time= 28 * 20= 560m .. so the answer is 560m.. l hope it helped :)
6 0
2 years ago
A baseball pitcher throws a ball across home plate. The ball travels 13.40m in 0.357s Determine the
Tanya [424]

Answer: I'm not sure what it needs to be rounded to, but I got 37.53501401 m/s

Explanation: The formula for speed is speed = distance/time. You plug in the distance (13.40) and the time (0.357), then divide 13.40 by 0.357

I hope this helps! :)

4 0
2 years ago
Name the four fundamental fores at work inside an atom. Tell what each one does.​
STALIN [3.7K]

Answer:

Four fundamental forces are gravitational, electromagnetic, strong, and weak.

Explanation:

The gravitational and electromagnetic interactions, which produce significant long-range forces whose effects can be seen directly in everyday life and the strong and weak interactions, which produce forces at minuscule, subatomic distances and govern nuclear interactions.

6 0
3 years ago
How do you convert seconds to minutes, seconds to hours , minutes to seconds and hours to seconds
Bess [88]

Answer:

You could memorize conversions, or use conversion charts, or do a quick internet search for help.

Explanation:

I'll do a few of the conversions for you

<u>Seconds to minutes: </u>

there are 60 seconds in one minute. So if you are wondering how many seconds are in 3 and 1/2 minutes, you would do this conversion:

60 sec/min <em>times </em>x sec/3.5 min

which can be written as 60 x 3.5 = 210. so "x" seconds would be 210 seconds in 3 1/2 minutes.

<u>you could do the same thing in the opposite direction for minutes to seconds:</u>

1 min has 60 seconds. So in 7.25 minutes, how many seconds are there?

1 min/60 sec <em>times </em>7.25 min/x sec

which can be written as 7.25 x 60 = 435. so "x" seconds would be 435 seconds in 7.25 minutes.

<u>hours to seconds:</u>

this one is slightly more complicated

In one hour there are 60 minutes, and in one minute there are 60 seconds.

so to convert from hours to seconds you would do this conversion:

1 hr/60 min times 1 min/60 sec. then the "min" would cancel out, and you would be left with the label "hr/sec". to do the math, it would be 1 hr / 60 x 60.

60x60 = 3600. so you would have 1 hr/3600 sec. So in one hour there are 3600 seconds.

so if you want to know how many seconds are in 6.75 hours:

6.75 hr/x sec <em>times </em>3600 sec/1 hr

6.75 x 3600 = 24,300 so there are 24,300 seconds in 6.75 hours.

I hope this helps :)

4 0
4 years ago
A vector → A has a magnitude of 56.0 m and points in a direction 30.0° below the negative x axis. A second vector, → B , has a m
MissTica

Answer:

  • The magnitude of the vector \vec{C} is 107.76 m

Explanation:

To find the components of the vectors we can use:

\vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | is the magnitude of the vector, and θ is the angle over the positive x axis.

The negative x axis is displaced 180 ° over the positive x axis, so, we can take:

\vec{A} = 56.0 \ m \ ( \ cos( 180 \° + 30 \°) \ , \ sin (180 \° + 30 \°) \ )

\vec{A} = 56.0 \ m \ ( \ cos( 210 \°) \ , \ sin (210 \°) \ )

\vec{A} = ( \ -48.497 \ m \ , \ - 28 \ m \ )

\vec{B} = 82.0 \ m \ ( \ cos( 180 \° - 49 \°) \ , \ sin (180 \° - 49 \°) \ )

\vec{B} = 82.0 \ m \ ( \ cos( 131 \°) \ , \ sin (131 \°) \ )

\vec{B} = ( \ -53.797 \ m \ , \ 61.886\ m \ )

Now, we can perform vector addition. Taking two vectors, the vector addition is performed:

(a_x,a_y) + (b_x,b_y) = (a_x+b_x,a_y+b_y)

So, for our vectors:

\vec{C} = ( \ -48.497 \ m \ , \ - 28 \ m \ ) + ( \ -53.797 \ m \ ,  ) = ( \ -48.497 \ m \ -53.797 \ m , \ - 28 \ m \ + \ 61.886\ m \ )

\vec{C} = ( \ - 102.294 \ m , \ 33.886 m \ )

To find the magnitude of this vector, we can use the Pythagorean Theorem

|\vec{C}| = \sqrt{C_x^2 + C_y^2}

|\vec{C}| = \sqrt{(- 102.294 \ m)^2 + (\ 33.886 m \)^2}

|\vec{C}| =107.76 m

And this is the magnitude we are looking for.

5 0
3 years ago
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