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zloy xaker [14]
3 years ago
5

A sumo wrestler is 330 pounds. How much force does the earth exert in him? Define Mass Define Weight

Physics
1 answer:
Oksana_A [137]3 years ago
5 0

Answer:

a large matters of body with no shape is mass

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Find the magnitude of a fourth force on the stone that will make the vector sum of the four forces zero.
Tanya [424]
100cos30° + 80cos120° + 40cos233° + Dx = 0 
<span>Dx = -22.53 N </span>

<span>y components: </span>
<span>100sin30° + 80sin120° + 40sin233° + Dy = 0 </span>
<span>Dy = -87.34 N </span>

<span>magnitude of D: </span>
<span>sqrt[(-22.53)² + (-87.34)²] </span>
<span>90.2 N </span>

<span>direction of D: </span>
<span>arctan[(-87.34)/(-22.53)] </span>
<span>75.5° ref, but since Dx and Dy are both negative, we know this vector is in QIV: </span>
<span>360 - 75.5° = 284.5°</span>
7 0
3 years ago
A mass attached to a spring is displaced from its equilibrium position by 5cm and released. The system then oscillates in simple
jok3333 [9.3K]

Answer:

C) 1 s

Explanation:

The period of a mass-spring system is given by the formula:

T=2\pi \sqrt{\frac{m}{k}}

where

m is the mass hanging on the spring

k is the spring constant

As we can see from the equation above, the period of the system does NOT depend on the initial amplitude of the oscillation. Therefore, even if the initial amplitude is changed from 5 cm to 10 cm, the period of the system will remain the same, 1 s.

4 0
4 years ago
(BRAINLIEST)
UkoKoshka [18]

Answer:

Gravity

because it's factorised by mass of a body.

For other forces, they deal with charges of negligible mass and weights

4 0
4 years ago
Read 2 more answers
a car with a mass of 2,000 kilograms is moving around a circular curve at a uniform velocity of 25 meters per second. the curve
Ber [7]
Fc=mv^2/r so we get 

2000kg*(25m/s)^2/(80m)= 15625N of force 

hope this helps! Thank You!!

4 0
3 years ago
7. Water flows trough a horizontal tube of diameter 2.5 cm that is joined to a second horizontal tube of diameter 1.2 cm. The pr
Usimov [2.4K]

To solve this problem it is necessary to apply the concepts related to the continuity of fluids in a pipeline and apply Bernoulli's balance on the given speeds.

Our values are given as

d_1 = 2.5cm \rightarrow r_1=1.25cm=1.25*10^{-2}m

d_2 = 1.2cm \rightarrow r_2 = 0.6cm = 0.6*10^{-2}m

From the continuity equations in pipes we have to

A_1V_1 = A_2 V_2

Where,

A_{1,2} = Cross sectional Area at each section

V_{1,2} = Flow Velocity at each section

Then replacing we have,

(\pi r_1^2) v_1 = (\pi r_2^2) v_2

(1.25*10^{-2})^2 v_1 = ( 0.06*10^{-2})^2 v_2

v_2 = \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1

From Bernoulli equation we have that the change in the pressure is

\Delta P = \frac{1}{2} \rho (v_2^2-v_1^2)

7.3*10^3 = \frac{1}{2} (1000)([ \frac{(1.25*10^{-2})^2 }{0.6*10^{-2})^2} v_1 ]^2-v_1^2)

7300= 8919.01 v_1^2

v_1 = 0.9m/s

Therefore the speed of flow in the first tube is 0.9m/s

6 0
4 years ago
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