Ask question

Ask question

All categories

3 years ago

10

A woman falls to the ground while wearing a parachute. The air resistance of a parachute is 500N. If the woman falls at a consta

nt rate of 5m/s, then the gravitational force on her is

1 answer:

Verizon [17]3 years ago

5 0

Answer:

500N

as in case of uniform motion net force acting is zero

You might be interested in

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of grav

erastovalidia [21]

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = \frac{7.4}{cos(70)}$

$Fm = 21.636N$

<em>For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that</em> $\beta = \theta$

3 0

3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

4 0

3 years ago

What is static friction?

givi [52]

Answer:

Plzzzzzzzzzzzzzzzz brainliest

Explanation:

In static friction, the frictional force resists force that is applied to an object, and the object remains at rest until the force of static friction is overcome. In kinetic friction, the frictional force resists the motion of an object. ... The frictional force itself is directed oppositely to the motion of the object.

7 0

2 years ago

If a pebble is being transported in a stream by rolling, how does the velocity of it compare to the velocity of the stream?

AleksandrR [38]

Streams carry sediment, like pebbles, in their flows. The pebbles can be in a variety of locations in the flow, depending on it's size, the balance between the upwards velocity on the pebble (drag and lift forces), and it's settling velocity.

3 0

3 years ago

A rowboat passenger uses an oar to push the boat off the dock by exerting a force of 40N for 3.0s. What impulse acts on the boat

jarptica [38.1K]

Answer:

120 Ns

Explanation:

The impulse exerted on an object is given by:

$I=Ft$

where

F is the force applied

t is the time taken

In this problem, we have:

F = 40 N

t = 3.0 s

So, the impulse acting on the boat is

$I=Ft=(40 N)(3.0 s)=120 Ns$

8 0

3 years ago