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pochemuha
3 years ago
7

How does the current change if the resistance is doubled?

Physics
1 answer:
finlep [7]3 years ago
7 0
The current will be divided by 2
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Three deer, a, b, and c, are grazing in a field. deer b is located 62.1 m from deer a at an angle of 54.3 ° north of west. deer
sveticcg [70]

by cosine law we know that

c^2 = a^2 + b^2 - 2 abcos\theta

\theta = 180 - 54.3 - 79.1 = 46.6 degree

now using above equation

93.8^2 = 62.1^2 + b^2 - 2*62.1*b * cos46.6

4942.03 = b^2 - 85.34 b

b^2 - 85.34b - 4942.03 = 0

by solving above quadratic equation we have

b = 124.9 m

so it is at distance 124.9 m from deer a

4 0
3 years ago
In trampoline competitions, a good jump is one that lasts about 1.8 seconds. (A) How high can an athlete who stays in the air 1.
KonstantinChe [14]

Answer:

3.97305 m

Explanation:

a = Acceleration due to gravity = 9.81 m/s²

If a jump lasts for 1.8 seconds this means that from the moment when the person leaves the ground till the person touches the ground again it takes 1.8 seconds. So, maximum height reached will be at half the time of the jump i.e., 0.9 seconds.

u = Initial velocity = 0

Equation of motion

s=ut+\frac{1}{2}at^2\\\Rightarrow s=0t+\frac{1}{2}9.81\times 0.9^2\\\Rightarrow s=3.97305\ m

So, height of the jump is 3.97305 m.

4 0
3 years ago
A 2.5g copper penny is given a charge of -4.0*10^-9c. how mny excess electrons are on the penny?
Lyrx [107]

Answer : The excess of electrons on the penny are, 2.5\times 10^{10} electrons

Solution : Given,

Total charge = -4.0\times 10^{-9}C

Charge on electron = -1.6\times 10^{-19}C

Formula used :

\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}

Now put all the given values in this formula, we get the excess of electrons present on the penny.

\text{Excess of electrons}=\frac{\text{Total charge}}{\text{Charge of electron}}=\frac{-4.0\times 10^{-9}C}{-1.6\times 10^{-19}C/e^-}=2.5\times 10^{10}e^-

Therefore, the excess of electrons on the penny are, 2.5\times 10^{10} electrons


7 0
3 years ago
in a certain pinhole camera, the screen is 10 cm from the pinhole. When the pinhole is placed 6 cm away from a tree ,a sharp ima
vagabundo [1.1K]

Answer:

Height of the tree is <u>9.6 cm</u>

Explanation:

We know that, Magnification of an image is written as follows.

\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)

Where,

\begin{array}{l}{\mathrm{H}_{0}=\text { height of the object }} \\ {\mathrm{H}_{\mathrm{i}}=\text { height of the image }} \\ {\mathrm{D}_{0}=\text { distance of the object }} \\ {\mathrm{D}_{\mathrm{i}}=\text { distance of the image }}\end{array}

As per given question,

\begin{array}{l}{\mathrm{H}_{1}=\text { height of the image }=\text { height of the image of the tree on screen }=16 \mathrm{cm}} \\ {\mathrm{D}_{0}=\text { distance of the object }=\text { distance of the tree from the pinhole }=6 \mathrm{cm}} \\ {D_{1}=\text { distance of the image }=\text { distance of the image from the pinhole }=10 \mathrm{cm}} \\ {\mathrm{H}_{0}=\text { height of the object }=\text { height of the tree }}\end{array}

Substitute the values in the above formula,

\begin{array}{l}{\left(\frac{H_{i}}{H_{o}}\right)=\left(\frac{D_{i}}{D_{o}}\right)} \\ {\left(\frac{16}{H_{o}}\right)=\left(\frac{10}{6}\right)} \\ {\mathrm{H}_{\mathrm{o}}=\left(\frac{16 \times 6}{10}\right)} \\ {\mathrm{H}_{\mathrm{o}}=9.6 \mathrm{cm}}\end{array}

Height of the tree is 9.6 cm.

8 0
3 years ago
How does an increase in cold working effect Modulus of Elasticity and why?
VladimirAG [237]

Answer:

There is a decrease in modulus of elasticity

Explanation:

Young's Modulus of elasticity also known as elastic modulus is the deformation of a body along a particular axis under the action of opposing forces along that axis. at atomic levels, it depends on bond energy or strength.

In cold working processes, plastic deformation a metal occurs below its re-crystallization temperature due to which crystal structure of metal gets distorted and as a result of dislocations fractures also occur resulting in hardening of metal but bonds at atomic levels defining elasticity are temporarily affected.

Thus an increase in cold working results in a decrease in modulus of elasticity.

7 0
3 years ago
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