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denis-greek [22]

3 years ago

13

Force of 10N down,10N to the right,and 5N to the left are acting on a ball .it acceleration horizontally to the right.what other

force,if any,is acting on the ball

1 answer:

denpristay [2]3 years ago

3 0

Answer:

A 10 N force pointing up

Explanation:

If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).

Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

Then there must also be a 10 N force pointing up acting on the object.

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In some circumstances, it is useful to look at the linear velocity of a point on the blade. The linear velocity of a point in un

mihalych1998 [28]

Answer:

$v=wr$

Explanation:

<u>Tangent and Angular Velocities</u>

In the uniform circular motion, an object describes the same angles in the same times. If $\theta$ is the angle formed by the trajectory of the object in a time t, then its angular velocity is

$\displaystyle w=\frac{\theta}{t}$

if $\theta$ is expressed in radians and t in seconds the units of w is rad/s. If the circular motion is uniform, the object forms an angle $2\theta$ in 2t, or $3\theta$ in 3t, etc. Thus the angular velocity is constant.

The magnitude of the tangent or linear velocity is computed as the ratio between the arc length and the time taken to travel that distance:

$\displaystyle v=\frac{\theta r}{t}$

Replacing the formula for w, we have

$\boxed{ v=wr}$

4 0

4 years ago

When a rocket is 4 kilometers high, it is moving vertically upward at a speed of 400 kilometers per hour. At that instant, how f

Y_Kistochka [10]

Answer:

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

Explanation:

Geometrically speaking, the distance between the rocket and the observer ( $r$ ), measured in kilometers, can be represented by a right triangle:

$r = \sqrt{x^{2}+y^{2}}$ (1)

Where:

$x$ - Horizontal distance between the rocket and the observer, measured in kilometers.

$y$ - Vertical distance between the rocket and the observer, measured in kilometers.

The angle of elevation of the rocket ( $\theta$ ), measured in sexagesimal degrees, is defined by the following trigonometric relation:

$\tan \theta = \frac{y}{x}$ (2)

If we know that $x = 5\,km$ , then the expression is:

$\tan \theta = \frac{y}{5}$

And the rate of change of this angle is determined by derivatives:

$\sec^{2}\theta \cdot \dot \theta = \frac{1}{5}\cdot \dot y$

$\frac{\dot \theta}{\cos^{2}\theta} = \frac{\dot y}{5}$

$\frac{\dot \theta\cdot (25+y^{2})}{25} = \frac{\dot y}{5}$

$\dot \theta = \frac{5\cdot \dot y}{25+y^{2}}$

Where:

$\dot \theta$ - Rate of change of the angle of elevation, measured in sexagesimal degrees.

$\dot y$ - Vertical speed of the rocket, measured in kilometers per hour.

If we know that $y = 4\,km$ and $\dot y = 400\,\frac{km}{h}$ , then the rate of change of the angle of elevation is:

$\dot \theta = 48.780\,\frac{\circ}{s}$

The angle of elevation of the rocket is increasing at a rate of 48.780º per second.

3 0

3 years ago

A single Oreo cookie provides 53 kcal of energy. An athlete does an exercise that involves repeatedly lifting (without accelerat

Sever21 [200]

Answer:

Approximately $325$ (rounded down,) assuming that $g = 9.81\; {\rm N \cdot kg^{-1}}$ .

The number of repetitions would increase if efficiency increases.

Explanation:

Ensure that all quantities involved are in standard units:

Energy from the cookie (should be in joules, ${\rm J}$ ):

$\begin{aligned} & 53\; {\rm kCal} \times \frac{1\; {\rm kJ}}{4.184\; {\rm kCal}} \times \frac{1000\; {\rm J}}{1\; {\rm kJ}} \approx 2.551 \times 10^{5}\; {\rm J} \end{aligned}$ .

Height of the weight (should be in meters, ${\rm m}$ ):

$\begin{aligned} h &= 2\; {\rm dm} \times \frac{1\; {\rm m}}{10\; {\rm dm}} = 0.2\; {\rm m}\end{aligned}$ .

Energy required to lift the weight by $\Delta h = 0.2\; {\rm m}$ without acceleration:

$\begin{aligned} W &= m\, g\, \Delta h \\ &= 100\; {\rm kg} \times 9.81\; {\rm N \cdot kg^{-1}} \times 0.2\; {\rm m} \\ &= 196\; {\rm N \cdot m} \\ &= 196\; {\rm J} \end{aligned}$ .

At an efficiency of $0.25$ , the actual amount of energy required to raise this weight to that height would be:

$\begin{aligned} \text{Energy Input} &= \frac{\text{Useful Work Output}}{\text{Efficiency}} \\ &= \frac{196\; {\rm J}}{0.25} \\ &=784\; {\rm J}\end{aligned}$ .

Divide $2.551 \times 10^{5}\; {\rm J}$ by $784\; {\rm J}$ to find the number of times this weight could be lifted up within that energy budget:

$\begin{aligned} \frac{2.551 \times 10^{5}\; {\rm J}}{784\; {\rm J}} &\approx 325 \end{aligned}$ .

Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.

4 0

2 years ago

Can someone tell me how this circuit works?

ch4aika [34]

Answer:

In the scientific model, electric current is the overall movement of charged particles in one direction. The cause of this movement is an energy source like a battery, which pushes the charged particles. The charged particles can move only when there is a complete conducting pathway (called a ‘circuit’ or ‘loop’) from one terminal of the battery to the other.

A simple electric circuit can consist of a battery (or other energy source), a light bulb (or other device that uses energy), and conducting wires that connect the two terminals of the battery to the two ends of the light bulb. In the scientific model for this kind of simple circuit, the moving charged particles, which are already present in the wires and in the light bulb filament, are electrons.

Electrons are negatively charged. The battery pushes the electrons in the circuit away from its negative terminal and pulls them towards the positive terminal (see the focus idea Electrostatics – a non contact force). Any individual electron only moves a short distance. (These ideas are further elaborated in the focus idea Making sense of voltage). While the actual direction of the electron movement is from the negative to the positive terminals of the battery, for historical reasons it is usual to describe the direction of the current as being from the positive to the negative terminal (the so-called ‘conventional current’).

The energy of a battery is stored as chemical energy (see the focus idea Energy transformations). When it is connected to a complete circuit, electrons move and energy is transferred from the battery to the components of the circuit. Most energy is transferred to the light globe (or other energy user) where it is transformed to heat and light or some other form of energy (such as sound in iPods). A very small amount is transformed into heat in the connecting wires.

The voltage of a battery tells us how much energy it provides to the circuit components. It also tells us something about how hard a battery pushes the electrons in a circuit: the greater the voltage, the greater is the push (see the focus idea Using energy).

Explanation:

6 0

3 years ago

if the separation distance between the moon and the planet is increased by a factor of 4 then the force gravitational is

erastova [34]

Answer:

Fg = (G * m1 * m2) / r^2

The gravitational force is Fg/16

Explanation:

Fg = (G * m1 * m2) / (4r)^2

Fg = (G * m1 * m2) / 16r

Therefore,

Fg / 16

8 0

3 years ago

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