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OleMash [197]
4 years ago
8

Two weights are hanging as shown in the figure .

Physics
1 answer:
Anna [14]4 years ago
6 0
Draw the free body diagram for the mass W2.

In it the only forces that appear are W2 (downward) and the Tension of the cable A (upward), TA

Net force = 0 => W2 = TA

Then TA = 200 N
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The current required to stimulate the heart during ventricular fibrillation is about 110mA (1000mA = 1A). Assuming that a conduc
algol [13]

Answer:

Explanation:

Given

current required I=110 mA

Internal Resistance R=300 \Omega

According to ohm law

Current flows in a conductor is directly Proportional to the voltage applied.

V\propto I

V=IR

V=110\times 10^{-3}\times 300

V=33 V  

5 0
3 years ago
The normal force of a parked car is 15,000 Newtons. The coefficient of static friction between the rubber of the tires and the a
Shtirlitz [24]

Answer:

11250 N

Explanation:

From the question given above, the following data were obtained:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

Friction and normal force are related by the following equation:

F = μR

Where:

F is the frictional force.

μ is the coefficient of static friction.

R is the normal force.

With the above formula, we can calculate the frictional force acting on the car as follow:

Normal force (R) = 15000 N

Coefficient of static friction (μ) = 0.75

Frictional force (F) =?

F = μR

F = 0.75 × 15000

F = 11250 N

Therefore, the frictional force acting on the car is 11250 N

3 0
3 years ago
How long does it take an automobile traveling 72.5 km/h to become even with a car that is traveling in another lane at 54.5 km/h
Novosadov [1.4K]

Answer:

7.83

Explanation:

Just using the speed ,distance and time relation..Let time be T

Use this formula and find the distance covered by automobile .

Do same for car ..

and at last subtract the distance of automobile to car =141.

i.e (Distance of automobile -Dist.of car )=141

4 0
4 years ago
Ysical Science A
lisov135 [29]
Answer:
4.5 Cm
Hope this help
3 0
3 years ago
Read 2 more answers
A small metal sphere has a mass of 0.14 g and a charge of -22.0 nc . it is 10 cm directly above an identical sphere with the sam
jonny [76]
For this problem, we use the Coulomb's law written in equation as:

F = kQ₁Q₂/d²
where
F is the electrical force
k is a constant equal to 9×10⁹ 
Q₁ and Q₂ are the charge of the two objects
d is the distance between the two objects

Substituting the values:

F = (9×10⁹)(-22×10⁻⁹ C)(-22×10⁻⁹ C)/(0.10 m)²
F = 0.0004356 N
6 0
4 years ago
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