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3 years ago

8

Based on the data given, what is the relationship between mass, gravitational force, speed at impact, and weight of the objects?

1 answer:

Nana76 [90]3 years ago

4 0

Answer:

gfhffgfhfgf

Explanation:

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Muscular Strength exercises focus on high______and low___

Advocard [28]

Answer:

d d d d dd d d d d d dd d d d d dd d

4 0

3 years ago

It is now 9:11 a.m. but when the bell rings at 9:12 a.m. Susie will be late for Mrs. Garner's U.S. History class for the 3rd tim

GaryK [48]

Answer:

3.1 m/s

Explanation:

The total distance she has to run is the addition of the three lengths:

47 + 63 + 76 = 186 meters.

She needs to cover it one minute (60 seconds). Therefore her speed must be:

186 m / 60 s = 3.1 m/s

6 0

3 years ago

How much pressure is applied to the ground

statuscvo [17]

Answer:

Pressure applied by the man= 285103.125 $Pa$ or 41.35 $lb/in^{2}$

Explanation:

Pressure is defined as the perpendicular force applied per unit area.

i.e. $Pressure=\frac{Force}{Area}$

Now, $Force= mg$

where, $m$ = mass of the body(man) = 93 kg

$g$ = acceleration due to gravity of Earth = 9.81 $m/{s^{2}}$

$Area$ covered is equal to the area of both stilts(a man generally stands on two feet)

therefore $Area=2(0.04)^{2}$ $m^{2}$

and putting in the values, we get,

$Pressure=\frac{93\times9.81}{2\times(0.04)^{2}}Nm^{-2}=285103.125Nm^{-2}$

Now we need to convert to our required units:

$1Nm^{-2}=1Pa\\1Pa=0.000145038lb/in^{2}$

(We can get the above result by individually converting kg to lb and meters to inches respectively)

Using the above relations we get,

$Pressure=285103.125Pa=0.000145038\times285103.125lb/in^{2}=41.35lb/in^{2}$

7 0

3 years ago

A 6.8 KG object moves with a velocity of 8 M/S what’s it's kinetic energy?

FromTheMoon [43]

Answer:

Explanation:

KE = ½mv² = ½(6.8)8² = 217.6 J

round as appropriate because that result is way too much precision for the inputs provided. Arguably should be 200 J based on the single significant digit of the velocity.

8 0

3 years ago

Cliff divers at Acapulco jump into the sea from a cliff 37.1 m high. At the level of the sea, a rock sticks out a horizontal dis

Stella [2.4K]

Answer:

$v_x = 4.87 m/s$

Explanation:

Height of the cliff is given as

$h = 37.1 m$

now the time taken by the diver to hit the surface is given as

$h = \frac{1}{2}gt^2$

$37.1 = \frac{1}{2}(9.8)t^2$

$t = 2.75 s$

Now in the same time it has to cover a distance of 13.39 m

so the speed in horizontal direction is given as

$v_x = \frac{x}{t}$

$v_x = \frac{13.39}{2.75}$

$v_x = 4.87 m/s$

3 0

4 years ago