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shtirl [24]
4 years ago
6

What resistance must be connected in parallel with a 633-Ω resistor to produce an equivalent resistance of 205 Ω?

Physics
1 answer:
alukav5142 [94]4 years ago
5 0

Answer:

303 Ω

Explanation:

Given

Represent the resistors with R1, R2 and RT

R1 = 633

RT = 205

Required

Determine R2

Since it's a parallel connection, it can be solved using.

1/Rt = 1/R1 + 1/R2

Substitute values for R1 and RT

1/205 = 1/633 + 1/R2

Collect Like Terms

1/R2 = 1/205 - 1/633

Take LCM

1/R2 = (633 - 205)/(205 * 633)

1/R2 = 428/129765

Take reciprocal of both sides

R2 = 129765/428

R2 = 303 --- approximated

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3 years ago
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Jain applies a input force of 100 N to a lever that has a Mechanical advantage of 3. What is the output force
Licemer1 [7]

Answer:

Output force = 300 N

Explanation:

Given that,

The mechanical advantage of a lever, m = 3

Input force, F_i=100\ N

We need to find the output force. The ratio of output force to the input force is called the mechanical advantage of the lever. So,

m=\dfrac{F_o}{F_i}\\\\F_o=m\times F_i\\\\F_o=3\times 100\\\\F_o=300\ N

So, the output force is 300 N.

8 0
3 years ago
The chart below summarizes the forces applied to four different objects.
posledela

Answer:

C. Y

Explanation:

From Newton's second law of motion, we know that:

       Force = mass x acceleration

So;

      acceleration  = \frac{Force }{mass}  

 Therefore, to have the highest acceleration at a constant force, the mass must be low. Acceleration is inversely proportional to mass.

 

Y has the least mass and it will have the highest acceleration

6 0
3 years ago
A projectile of mass 9.6 kg is launched from the ground with an initial velocity of 12.4 m/s at angle of 54° above the horizonta
Temka [501]

Answer:

The location is at (3.535, 1.162) m

Solution:

As per the question:

Mass of the projectile, m = 9.6 kg

Initial velocity, v = 12.4 m/s

Angle, \theta = 54^{\circ}

Mass of one fragment, m = 6.5 kg

Time taken by the fragment, t = 1.42 s

Height of the fragment, y = 5.9 m

Horizontal distance, x = 13.6 m

Now,

To determine the location of the second fragment:

Horizontal Range, R = \frac{v^{2}sin2\theta}{g}

R = \frac{12.4^{2}sin2(54)}{9.8} = 14.92\ m

Time of flight, t' = \frac{2vsin\theta}{g} = \frac{2\times 12.4sin108}{9.8}= 2.406\ s

Now, for the fragments:

Mass of the other fragment, m' = M - m = 9.6 - 6.5 = 3.1 kg

Distance traveled horizontally:

s_{x} = vcos\theta = 12.4cos54^{\circ}\times 1.42 = 10.35\ m

Distance traveled vertically:

s_{y} = vcos\theta - \frac{1}{2}gt^{2}

s_{y} = 12.4sin54^{\circ}\times 1.42 -  \frac{1}{2}\times 9.8\times 1.42^{2} = 14.25 - 9.88 = 4.37\ m

Now,

s_{x} = \frac{mx + m'x'}{M}

10.35= \frac{6.5\times 13.6 + 3.1x'}{9.6}

x' = 3.535 m

Similarly,

s_{y} = \frac{my + m'y'}{M}

4.37= \frac{6.5\times 5.9 + 3.1y'}{9.6} = 1.162\ m

The location of the other fragment is at (3.535, 1.162)

5 0
4 years ago
How do I determine which object has the greatest acceleration?
IceJOKER [234]


To determine which object has the greatest acceleration, you would use the formula:

acceleration = (Final Velocity - Initial Velocity) / Time

Just plug-in values for what you have and see which one has the greatest acceleration. If you're given velocity-time graphs, then the one with the steepest slope would have the greatest acceleration.

8 0
3 years ago
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