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3 years ago

How do you solve c=8.50 m at 90 degrees

1 answer:

8 0

Step #1):
You tell us what 'c' means.

Step #2):
You tell us whether the 90 degrees is an angle or a temperature.

Step #3):
You tell us the relationship between 'c' and the angle or the temperature.

Step #4):
We help you select the appropriate laws of Physics and apply them
to the situation.

You might be interested in

That 1.5 kg brick falls to the ground. What is the kinetic energy when the brick is moving at 26 m/s? Formula: Variables: Work w

Tresset [83]

Answer:

Explanation:

The kinetic energy of an object can be found by using the formula

$k = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$k = \frac{1}{2} \times 1.5 \times {26}^{2} \\ = \frac{1}{2} \times 1.5 \times 676 \\ = 1.5 \times 338 \\ = 507$

We have the final answer as

Hope this helps you

5 0

3 years ago

A bike travels 4 miles in half an hour, what is its speed?

omeli [17]

Answer:

8 mph

Explanation:

4 miles in half hour so you add 4 more for the second half

3 0

3 years ago

WILL GIVE BRAINLIEST!!!!!!!!!!!!!!!!!!

faust18 [17]

Answer:

9)a

10) I think true

11)b

Explanation:

9)a. because it's told that the car is slowing down, the sum of the forces that are towards left, should be more than the ones that are towards right. if the car was gaining speed, "b" would have been correct. and if it was told that the car is moving without a change in the speed, "c" would have been correct.

10) if a moving object has a change of speed or direction, it would have an acceleration. now if a moving object experiences an unbalanced force, it'd either slow down, gain speed or change direction, and in all of the three possibilities it'd have an acceleration.

11) upward and downward forces are equal, and the sum of them would be 0N(because they have opposite directions). so they negate each other.

and the rightward force is 5N more than the leftward force. so the Net Force would be 5N.

-30+30-10+15=5N

if it is unclear or you need more explanation, ask freely.

8 0

3 years ago

Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has

faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

Charge on first charged particle, $q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.$
Charge on the second charged particle, $q_2=3.80\ nC=3.80\times 10^{-9}\ C.$

Position of the first charge = $(x_1=0.00\ m,\ y_1=0.600\ m).$
Position of the second charge = $(x_2=1.50\ m,\ y_2=0.650\ m).$

The electric field at a point due to a charge $q$ at a point $r$ distance away is given by

$\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.$

where,

$k$ = Coulomb's constant, having value $\rm 8.99\times 10^9\ Nm^2/C^2.$

$\vec r$ = position vector of the point where the electric field is to be found with respect to the position of the charge $q$ .

$\hat r$ = unit vector along $\vec r$ .

The electric field at the origin due to first charge is given by

$\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.$

$\vec r_1$ is the position vector of the origin with respect to the position of the first charge.

Assuming, $\hat i,\ \hat j$ are the units vectors along x and y axes respectively.

$\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.$

Using these values,

$\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.$

The electric field at the origin due to the second charge is given by

$\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.$

$\vec r_2$ is the position vector of the origin with respect to the position of the second charge.

$\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.$

Using these values,

$\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\ N/C.$

The net electric field at the origin due to both the charges is given by

$\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.$