Muscular Strength exercises focus on high______and low__

Which kind of force do you exert on an object when you pull it toward you?

jarptica [38.1K]

By definition we have to:

Applied force: It is the external force that acts directly on a body.

Therefore, we can say that if you have an object and push it towards yourself, you are exerting an external force on the object.

This external force was not acting on the object previously, therefore, it is a force that you are applying at that moment.

Answer:

you exert an Applied Force on an object when you pull it towards you

A. Applied Force

4 0

3 years ago

Read 2 more answers

E)

guapka [62]

Answer:

Pressure, P = 32666.66 Pa

Explanation:

It is given that,

Surface area of foot of Bimaba is 150 cm² or 0.015 m².

Her weight is 50 kg

We need to find the pressure does she exert on the ground, as she stands on her one foot. The force acting per unit area is called pressure. It can be given by :

$P=\dfrac{F}{A}\\\\P=\dfrac{mg}{A}\\\\P=\dfrac{50\times 9.8}{0.015}\\\\P=32666.66\ Pa$

So, the pressure is 32666.66 Pa.

7 0

3 years ago

Which statement best describes the energy changes that occur while a child is riding on a sled down a steep, snow-covered hill?

Degger [83]

B. kinetic energy increases and potential energy decreases

5 0

3 years ago

A stunt driver wants to make his car jump over 8 cars parked side by side below a horizontal ramp.

Genrish500 [490]

<u>Answer:</u>

a) Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

<u>Explanation:</u>

a) The height of ramp = 1.5 meter

Horizontal distance he must clear = 22 meter

The car is having horizontal motion and vertical motion. In case of vertical motion the acceleration on the car is acceleration due to gravity.

We have equation of motion, $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In case of vertical motion initial velocity = 0 m/s, acceleration = 9.8 $m/s^2$ , we need to calculate time when displacement = 1.5 meter.

$1.5=0*t+\frac{1}{2} *9.8*t^2\\ \\ t = 0.553 seconds$

So the car has to cover a distance of 22 meter in 2.119 seconds.

So minimum speed required = 22/0.553 = 39.78 m/s

Minimum speed must he drive off the horizontal ramp = 39.78 m/s

b) When the take of angle is 7⁰ the vertical speed of car is not zero = V sin 7 = 0.122 V

So the in case of vertical motion we have initial velocity = 0.122 V, S = -1.5 meter( below ramp), acceleration = -9.8 $m/s^2$

Substituting

$-1.5=0.122V*t-\frac{1}{2} *9.8*t^2\\ \\ 4.9t^2-0.122Vt-1.5=0$

In case of horizontal motion

Horizontal speed of car = V cos 7 = 0.993V

So it has to travel 22 meter in t seconds

0.993Vt = 22, Vt = 22.155 m

Substituting in the equation $4.9t^2-0.122Vt-1.5=0$

We will get $4.9t^2-0.122*22.155-1.5=0\\ \\ t = 0.926 seconds$

Speed required = 22.155/0.926 = 23.93 m/s

Minimum speed must he drive off the horizontal ramp with 7° above the horizontal = 23.93 m/s

7 0

3 years ago

When helping someone whose car has a "dead" battery, how should your car's battery be connected in relation to the dead battery?

leonid [27]

Positive terminal with Positive terminal. The negative terminal should be connected to the car chassis.

8 0

3 years ago

Muscular Strength exercises focus on high______and low___

Muscular Strength exercises focus on high____and low_