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2 years ago

Plz answer all 5 of the questions i will give brainlyest

Physics

1 answer:

lubasha [3.4K]2 years ago

6 0

Answer:

1)force:- force is an external affert by any body on an object.

2) force and motion are related to each other. force helps to create motion in any object.

3) frictional force

4) Newton

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When waves strike an object and bounce off, what has occured

Vesna [10]

The answer is reflection

5 0

2 years ago

A straight line is drawn on the surface of a 14-cm-radius turntable from the center to the perimeter. A bug crawls along this li

sleet_krkn [62]

Answer:

v = $\left[\begin{array}{c}0.66&0\end{array}\right]$ m/s

Explanation:

The position vector r of the bug with linear velocity v and angular velocity ω in the laboratory frame is given by:

$\overrightarrow{r}=vtcos(\omega t)\hat{x}+vtsin(\omega t)\hat{y}$

The velocity vector v is the first derivative of the position vector r with respect to time:

$\overrightarrow{v}=[vcos(\omega t)-\omega vtsin(\omega t)]\hat{x}+[vsin(\omega t)+\omega vtcos(\omega t)]\hat{y}$

The given values are:

$t=\frac{x}{v}=\frac{14}{3.8}=3.7 s$

$\omega=\frac{45\times2\pi}{60s}=4.7\frac{1}{s}$

8 0

2 years ago

An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10

STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

$P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}$

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi $(\frac{6894.76\ Pa}{1\ psi})$ = 379212 Pa

$\rho$ = density of water = 1000 kg/m³

Therefore,

$v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}$

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

$\frac{V}{t} = Av\\\\t =\frac{V}{Av}$

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = $\frac{\pi (0.015875\ m)^2}{4}$ = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = $[\frac{\pi (9.144\ m)^2}{4}][1.524\ m]$ = 100.1 m³

Therefore,

$t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\$

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0

2 years ago

Determine a formula for the acceleration of the system in terms of mA, mB, θ, and g. Ignore the mass of the cord and pulley. Exp

jekas [21]

Answer:

$a=\frac{mBg-mAgSin\theta}{mA+mB}$

Explanation:

Given two mass on an incline code $mA$ and $mB$ and an angle of inclination $\theta$ . $g$ . Assume that $mA$ is the weight being pulled up and $mB$ the hanging weight.

-The equations of motion from Newton's Second Law are:

$mBg-T=mBa$ where a is the acceleration.

#Substituting for $T$ (tension) gives:

$mBg-mAsin\theta-mAa=mBa$

#and solving for $a:$

$a=\frac{mBg-mAgSin\theta}{mA+mB}$ which is the system's acceleration.

8 0

3 years ago

61. A physics student has a single-occupancy dorm room. The student has a small refrigerator that runs with a current of 3.00 A

Mademuasel [1]

Answer:

Part a)

$percentage = 21.3$ %

Part b)

$percentage = 2.13 \times 10^{-5}$ %

Explanation:

As we know that total power used in the room is given as

$P = P_1 + P_2 + P_3 + P_4$

here we have

$P_1 = (110)(3) = 330 W$

$P_2 = 100 W$

$P_3 = 60 W$

$P_4 = 3 W$

$P = 330 + 100 + 60 + 3$

$P = 493 W$

Part a)

Since power supply is at 110 Volt so the current obtained from this supply is given as

$110\times i = 493$

$i = 4.48 A$

now resistance of transmission line

$R = \frac{\rho L}{A}$

$R = \frac{(2.8 \times 10^{-8})(10\times 10^3)}{\pi(4.126\times 10^{-3})^2}$

$R = 5.23 \ohm$

now power loss in line is given as

$P = i^2 R$

$P = (4.48)^2(5.23)$

$P = 105 W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{105}{493} \times 100$

$percentage = 21.3$ %

Part b)

now same power must have been supplied from the supply station at 110 kV, so we have

$110 \times 10^3 (i ) = 493$

$i = 4.48\times 10^{-3} A$

now power loss in line is given as

$P = i^2 R$

$P = (4.48 \times 10^{-3})^2(5.23)$

$P = 1.05 \times 10^{-4} W$

Now percentage loss is given as

$percentage = \frac{loss}{supply} \times 100$

$percentage = \frac{1.05 \times 10^{-4}}{493} \times 100$

$percentage = 2.13 \times 10^{-5}$ %

6 0

2 years ago