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ANEK [815]
3 years ago
14

a concrete slab 20 m long and weighing 400,000 N is supported by one pillar. A 19,600 N car is parked 8 meters from one end, whe

re should the pillar be so the slab is balanced?
Physics
1 answer:
Elden [556K]3 years ago
3 0

let the distance of pillar is "r" from one end of the slab

So here net torque must be balance with respect to pillar to be in balanced state

So here we will have

Mg(r - L/2) = mg(L/2 - 8)

here we know that

mg = 19600 N

Mg = 400,000 N

L = 20 m

from above equation we have

400,000(r - 10) = 19,600 (10 - 8)

r - 10 = 0.098

r = 10.098 m

so pillar is at distance 10.098 m from one end of the slab

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The answer would be “B” because humans would need water, protection from radiation so we don’t melt or burn to death lol, and a gaseous atmosphere because we would need oxygen.
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3 years ago
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lonnie pitches a baseball of mass 0.02kg. The ball arrives at home plate with a speed of 40 m/s and is batted straight back to L
vodomira [7]

Answer:

I=2 kg.m/s

Explanation:

The impulse is defined as the change of momentum:

I=p_f-p_o\\I=m*v_f-m*v_o\\I=0.02kg*[(-60m/s)-40m/s]\\I=2kg.m/s

We took the final velocity as negative since it is going on the opposite direction of the intial motion of the ball.

8 0
3 years ago
What is a type of pulley that increases the size and effort of force
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The answer is Movable
5 0
4 years ago
A 36.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130
konstantin123 [22]

Answer:

(a) W = 650J

(b) Wf = 529.2J

(c) W = 0J

(d) W = 0J

(e) ΔK = 120.8J

(f) v2 = 2.58 m/s

Explanation:

(a) In order to find the work done by the applied force you use the following formula:

W=Fd      (1)

F: applied force = 130N

d: distance = 5.0m

W=(130N)(5.0m)=650J

The work done by the applied force is 650J

(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:

W_f=F_fd=\mu Mgd       (2)

μ: coefficient of friction = 0.300

M: mass of the box = 36.0kg

g: gravitational constant = 9.8 m/s^2

W_f=(0.300)(36.0kg)(9.8m/s^2)(5.0m)=529.2J

The increase in the internal energy is 529.2J

(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.

W = 0J

(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.

(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:

\Delta K=W_T         (3)

You calculate the total work:

W_T=Fd-F_fd=(F-F_f)d     (4)

F: applied force = 130N

Ff: friction force

d: distance = 5.00m

The friction force is:

F_f=(0.300)(36.0kg)(9.8m/s^2)=105.84N

Next, you replace the values of all parameters in the equation (4):

W_T=(130N-105.84N)(5.00m)=120.80J

\Delta K=120.80J

The change in the kinetic energy of the box is 120.8J

(e) The final speed of the box is calculated by using the equation (3):

W_T=\frac{1}{2}M(v_2^2-v_1^2)       (5)

v2: final speed of the box

v1: initial speed of the box = 0 m/s

You solve the equation (5) for v2:

v_2 = \sqrt{\frac{2W_T}{M}}=\sqrt{\frac{2(120.8J)}{36.0kg}}=2.58\frac{m}{s}

The final speed of the box is 2.58m/s

5 0
3 years ago
Calculate the difference in blood pressure between the feet and top of the head for a person who is 1.70 m tall.
cupoosta [38]

Answer:

P_2 - P_1 = 1.8 * 10^4\ Pa

Explanation:

Given

Height (h) = 1.70m

Required

Determine the difference in the blood pressure from feet to top

This is calculated using Pascal's second law.

The second law is represented as:

P_2 = P_1 + pgd

Subtract P1 from both sides

P_2 - P_1 = pgd

Where

p = blood\ density = 1.06 * 10^3kg/m^3

g = acceleration\ of\ gravity = 9.8N/kg

d =height = 1.70m

P2 - P1 = Blood Pressure Difference

So, the expression becomes:

P_2 - P_1 = 1.06 * 10^3 * 9.8 * 1.70

P_2 - P_1 = 17659.6Pa

P_2 - P_1 = 1.8 * 10^4\ Pa

Hence, the difference in blood pressure is approximately 1.8 * 10^4\ Pa

3 0
3 years ago
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