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2 years ago

Alyssa is taking a walk on a nature path near her neighborhood. She

Physics

1 answer:

Montano1993 [528]2 years ago

4 0

Answer:

C. Sources of electrical current!

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Two blocks of masses m and M are connected by a string and pass over a frictionless pulley. Mass m hangs vertically, and mass M

horsena [70]

Answer:

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

$sin\theta - \mu_k cos\theta = 1$

Explanation:

Force of friction on M mass so that it will move down the inclined plane is given as

$F_f = \mu Mgcos\theta$

now if it is moving down the inclined plane at constant speed

so we will have

$Mgsin\theta - T - \mu mgcos\theta = 0$

on other side the mass "m" will go up at constant speed

so we have

$T - mg = 0$

so we have

$Mgsin\theta = \mu Mgcos\theta + mg$

so we have

$sin\theta - \mu_k cos\theta = \frac{m}{M}$

for special case when m = M

then we have

$sin\theta - \mu_k cos\theta = 1$

5 0

2 years ago

A convex mirror has a focal length of -10.8 cm. An object is placed 32.7 cm from the mirror's surface. Determine the image dista

KonstantinChe [14]

Answer:

-353.16

Explanation:

4 0

2 years ago

A leaky faucet drips once each second. at this rate, it takes 1 hour to fill a 1-liter graduated cylinder to the 200-cm3 mark. w

UNO [17]

Define
v = volume of a drop per second, cm³/s

The time taken to fill 200 cm³ is 1 hour.
Let V = 200 cm³, the filled volume.
Let t = 1 h = 3600 s, the time required to fill the volume.

Therefore,
$\frac{200 \, cm^{3}}{v \, cm^{3}/s} = 3600 \, s \\ v = \frac{200}{3600} =0.0556 \, cm^{3}$

The average volume of a single drop is approximately 0.0556 cm³.

Answer: 0.0556 cm³

8 0

2 years ago

A convex spherical mirror having a radius of curvature 18 cm (focal length = 1/2 radius of curvature for a spherical mirror) pro

tekilochka [14]

Answer:

distance between object and image = 18.9 cm

Explanation:

given data

radius of curvature = 18 cm

focal length = 1/2 radius of curvature

magnification = 40%

to find out

distance between object and image

solution

we know lens formula that is

1/f = 1/v + 1/u ....................1

here f = 18 /2 and v and u is object and image distance

and we know m = 40% = 0.40

so 0.40 = -v / u

so here v = - 0.40 u

so from equation 1

1/f = 1/v + 1/u

2/18 = - 1/0.40u + 1/u

u = -13.5 cm ..................2

and

v = -0.40 (- 13.5)

v = 5.4 cm ......................3

so from equation 2 and 3

distance between object and image = 5.4 + 13.5

distance between object and image = 18.9 cm

6 0

3 years ago

A rocket is launched at an angle of 53.0° above the horizontal with an initial speed of 103 m/s. The rocket moves for 3.00 s alo

Serggg [28]

Before the engines fail $(0\le t\le3.00\,\rm s)$ , the rocket's horizontal and vertical position in the air are

$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

$y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t^2$

and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

$v_y=\left(103\,\frac{\rm m}{\rm s}\right)\sin53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\sin53.0^\circ t$

After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}$

After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

$v_y=159\,\frac{\rm m}{\rm s}-gt$

where we take $g=9.80\,\frac{\rm m}{\mathrm s^2}$ .

a. The maximum altitude occurs at the point during which $v_y=0$ :

$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

b. The rocket will eventually fall to the ground at some point after its engines fail. We solve $y=0$ for $t$ , then add 3 seconds to this time:

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

$273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)(34.6\,\rm s)=4410\,\rm m$

5 0

2 years ago