Why your body does not tolerate vertical g-forces,

PLease help on this one and show work for brainliest?

Amiraneli [1.4K]

c .... work is force x distance as a definition

3 0

3 years ago

A 2 kg ball of putty moving to the right at 3 m/s has a perfectly inelastic, head-on collision with a 1 kg ball of putty moving

Aneli [31]

Answer:

$V=1.33m/s$ to the right

Explanation:

The balls collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V.

We need to use the conservation of momentum Law, the total momentum is the same before and after the collision.

In the axis X:

$m_{1}*v_{o1}-m_{2}*v_{o2}=(m_{1}+m_{2})V$ (1)

$V=(m_{1}*v_{o1}-m_{2}*v_{o2})/(m_{1}+m_{2})=(2*3-1*2)/(2+1)=1.33m/s$

5 0

2 years ago

A projectile is fired at an upward angle of 45.0º from the top of a 265-m cliff with a speed of .sm 185 What will be its speed w

nignag [31]

The final velocity of the projectile when it strikes the ground below is 198.51 m/s.

<h3>Time of motion of the projectile</h3>

The time taken for the projectile to fall to the ground is calculated as follows;

h = vt + ¹/₂gt²

where;

h is height of the cliff
v is velocity
t is time of motion

265 = (185 x sin45)t + (0.5)(9.8)t²

265 = 130.8t + 4.9t²

4.9t² + 130.8t - 265 = 0

solve the quadratic equation using formula method,

t = 1.89 s

<h3>Final velocity of the projectile</h3>

vyf = vyi + gt

where;

vyf is the final vertical velocity
vyi is initial vertical velocity

vyf = (185 x sin45) + (9.8 x 1.89)

vyf = 149.322 m/s

vxf = vxi

where;

vxf is the final horizontal velocity
vxi is the initial horizontal velocity

vxf = 185 x cos(45)

vxf = 130.8 m/s

vf = √(vyf² + vxf²)

where;

vf is the speed of the projectile when it strikes the ground below

vf = √(149.322² + 130.8²)

vf = 198.51 m/s

Learn more about final velocity here: brainly.com/question/6504879

#SPJ1

3 0

2 years ago

EM Waves have crests, troughs, and wavelengths <br> True or false?

Nezavi [6.7K]

True.........................................

4 0

3 years ago

Read 2 more answers

An object is dropped from an altitude of one Earth radiusabove Earth's surface. If M is the mass of Earth and R is itsradius the

Montano1993 [528]

Answer:

The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$

(A) is correct option.

Explanation:

Given that,

M = mass of earth

R = radius of earth

The potential energy at height above the surface of the earth

$P.E=-\dfrac{GmM}{R+h}$

The kinetic energy at height above the surface of the earth

$K.E = 0$

The total energy at height above the surface of the earth

$E = K.E+P.E$

$E = -\dfrac{GmM}{R+h}$ ....(I)

The total energy at the surface of the earth

$E'=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$ ....(II)

We need to calculate the speed of the object just before it hits Earth

From equation (I) and (II)

$-\dfrac{GmM}{R+h}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

Here, h = R

$-\dfrac{GmM}{2R}=\dfrac{1}{2}mv^2-\dfrac{GmM}{R}$

$v= \sqrt{\dfrac{GM}{R}}$

Hence, The speed of the object just before it hits Earth is $\sqrt{\dfrac{GM}{R}}$ .

4 0

3 years ago