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2 years ago

Find the voltage across the 15 Q resistor. [?] V No links please

Physics

1 answer:

Margaret [11]2 years ago

4 0

Answer:

Explanation:

same idea as before Liam, first, find the parallel resistance in 35 || 20

(35*20) / (35+20) = 700 / 55 = 12.727272 ohms

now add the 12.727272 + 15 = 27.727272 ohms total resistance

V = IR

10 = I * 27.727272

10 / 27.727272 = I

0.360655 = I

V = IR (again, but across the 15 ohm resistor)

V = 0.360655 * 15

V = 5.4098

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The acceleration due to gravity on Mars is less than that on Earth. On Mars, a person will weigh than on Earth.

pantera1 [17]

On mars people would way less.
An example of this is that if I weighed 700 pounds (I don't by the way) I would then weigh 500 pounds or less.

4 0

3 years ago

Read 2 more answers

Two ice skaters, Lilly and John, face each other while at rest, and then push against each other's hands. The mass of John is tw

seropon [69]

Answer:

Lilly's speed is two times John's speed.

Explanation:

m = Mass

a = Acceleration

t = Time taken

u = Initial velocity

v = Final velocity

The force they apply on each other will be equal

$F=ma\\\Rightarrow a_l=\frac{F}{m_l}$

$F=ma\\\Rightarrow a_j=\frac{F}{2m_l}\\\Rightarrow a_j=\frac{1}{2}a_l$

$v=u+at\\\Rightarrow v_l=0+\frac{F}{m_l}\times t\\\Rightarrow v_l=a_lt$

$v=u+at\\\Rightarrow v_l=0+\frac{F}{2m_l}\times t\\\Rightarrow v_j=\frac{1}{2}a_lt\\\Rightarrow v_j=\frac{1}{2}v_l\\\Rightarrow v_l=2v_j$

Hence, Lilly's speed is two times John's speed.

4 0

3 years ago

Assume it takes 10 J to stretch a spring 10 cm beyond its natural length. Find the work required (in Joules) to stretch the spri

Lady bird [3.3K]

Answer:

W = 30 J

Explanation:

given,

Work done = 10 J

Stretch of spring, x = 0.1 m

We know,

dW = F .dx

we know, F = k x

$\int dW = \int_0^{0.1} k.x dx$

$W = \int_0^{0.1} k.x dx$

$W = k[\dfrac{x^2}{2}]_0^{0.1}$

$10 = k\dfrac{0.1^2}{2}$

k = 2000

now, calculating Work done by the spring when it stretched to 0.2 m from 0.1 m.

$W = \int_{0.1}^{0.2} 2000 x dx$

$W = 2000 [\dfrac{x^2}{2}]_{0.1}^{0.2} dx$

W = 1000 x 0.03

W = 30 J

Hence, work done is equal to 30 J.

4 0

3 years ago

A space probe is coasting through space at a steady speed of 100p feet per second. the booster rocket fires for 1/2 seconds so t

aleksandr82 [10.1K]

Answer:

Acceleration: $9800 ft/s^2$

Explanation:

The acceleration of an object is equal to the rate of change of velocity:

$a=\frac{v-u}{t}$

where

u is the initial velocity

v is the final velocity

t is the time taken for the velocity to change from u to v

For the space probe in this problem, we have:

u = 100 ft/s (initial velocity)

v = 5000 ft/s (final velocity)

t = 0.5 s (time taken)

Therefore, the acceleration is

$a=\frac{5000-100}{0.5}=9800 ft/s^2$

6 0

3 years ago

A candle is 49 cm in front of a convex spherical mirror that has a focal length

vitfil [10]

The convex spherical mirror is the diverging mirror. The image distance is 20.4 cm. The magnification is 0.417. The image is virtual and upright.

<h3>What is magnification?</h3>

Magnification is the ratio of image distance to the object distance.

Object distance is 49 cm and the focal length of convex mirror is 35 cm, then image distance is calculated by the lens maker formula.

1/f = 1/v +1/u

1/-35 = 1`/v +1/49

v= -20.4 cm

Thus, the image distance is 20.4 cm.

The magnification is given by

m = v/u

m = -20.4 / 49

m = 0.417

Thus, the magnification is 0.417.

The image formed is by virtual meeting of light rays and above the principle axis. Thus the image is virtual and upright.

Learn more about magnification.

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5 0

2 years ago