What differentiates the isotopes of an element

how much force ks required.to.keep 780 n box moving at constant velocityu across.the floor if friction force betwern the bix and

icang [17]

Answer:

Explanation:you must first make a free body diagram of the block to place all the forces that act on the block there. You will have four forces. The normal force, the weight, the horizontal force in x (which is what they ask to calculate) and the friction force. Then you raise the equations and you clear the values. I attach the solution.

8 0

3 years ago

What must be the magnitude of a uniform electric field if it is to have the same energy density as that possessed by a 0.50 T ma

Sindrei [870]

Answer:

E = 1.50 × $10^{8}$ V/m

Explanation:

given data

B = 0.50 T

solution

we know that energy density by the magnetic field is express as

$\mu _b = \frac{B}{2\mu _o}$ ...............1

and

energy density due to electric filed is

$\mu _e = \frac{\epsilon _o E^2}{2}$ ...............2

and here $\mu _b = \mu _ e$

so that

E = $\frac{B}{\sqrt{\mu _o \times \epsilon _o}}$ ...................3

put here value and we get

$E = \frac{0.50}{\sqrt{4\pi \times 10^{-7} \times 8.852 \times 10^{-12}}}$

E = 3 × $10^{8}$ × 0.50

E = 1.50 × $10^{8}$ V/m

6 0

3 years ago

Which statement best describes the work and energy in these examples?

solniwko [45]

Answer: <u>Option A: </u>The gas and food are examples of energy.

Explanation:

Work and energy are inter-related. Energy is required to do the work. An equal amount of energy is converted into work. Work and energy have same units. The SI unit is Joule.

The gas and food are sources of energy. They act as fuel. This energy is utilized to perform work. The gas is used to run the car. The food is metabolized inside the body which is a source of energy and utilized to perform every day work.

Hence, The gas and food are examples of energy.

8 0

3 years ago

Read 2 more answers

A grinding wheel 0.35 m in diameter rotates at 2600 rpm .

ddd [48]

It is calculated that a)The angular velocity of the wheel is 272.13 rad/s,

b)On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

So, the radius, r = 0.35/2 = 0.175 m

As 1 revolution = 2π rad

(a)the angular velocity, ω = 2600 rpm = $\frac{2600 * 2\pi }{60}$ rad/s

⇒ω = 272.13 rad/s

So, the angular velocity is 272.13 rad/s.

(b)The linear speed, v = r * ω

⇒v = 0.175 * 272.13

⇒v= 47.62 m/s

(c)The angular acceleration, $a=\frac{v^{2} }{r}$

$a = \frac{(47.62)^{2} }{0.175}$

⇒ $a$ = 12958.08 m/s²

Learn more about angular velocity here:

brainly.com/question/13649539

#SPJ1

6 0

2 years ago

A cyclist makes the following trip along two vectors; he travels 9km to the north and then travels 6km to the east

elena-14-01-66 [18.8K]

Answer:

Final distance from the origin: 10.82 km. the vector points as shown in the attached image.

Angle with respect to the east: $56.31^o$

Explanation:

Please refer to the attached image. The cyclist's trip is indicated with the green arrows (9 km to the north followed by 6 km to the east.

So his final position is at the tip of this last vector, and indicated by the orange vector drawn form the point where the trip starts to the cyclist's final location.

We observe that this orange vector is in fact the hypotenuse of a right angle triangle, and we can estimate the distance from the origin by the Pythagorean theorem:

$d=\sqrt{9^2+6^2} \\d=\sqrt{81+36} \\d=\sqrt{117} \\d=10.82 \,\,km$

Notice that this is NOT the actual number of km that the cyclist pedaled to reach the final point.

Now, to find the value of the angle $\theta$ , we need to use trigonometry, and in particular the tangent function gives us the ratio between the side of the triangle "opposite" to the angle, divided the side "adjacent" to the angle:

$tan(\theta)=\frac{opp}{adj} \\tan(\theta)=\frac{9}{6}\\tan(\theta)=\frac{3}{2}\\$

Now we can find the value of the angle by using the arctan function:

$tan(\theta)=\frac{3}{2} \\\theta=arctan(\frac{3}{2} )\\\theta= 56.31^o$

6 0

3 years ago