Which of the following confirms that gravitational mass and inertial mass are equivalent?

A spherical shell contains three charged objects. The first and second objects have a charge of − 14.0 nC and 33.0 nC , respecti

matrenka [14]

Explanation:

Formula depicting relation between total flux and total charge Q is as follows.

$\phi = \frac{Q}{\epsilon_{o}}$ (Gauss's Law)

Putting the given values into the above formula as follows.

Q = $\phi \times \epsilon_{o}$

= $-953 Nm^{2}/C \times 8.854 \times 10^{-12}$

= $-8.4 \times 10^{-9} C$

= -8.4 nC

Therefore, when the unknown charge is q then,

-14.0 nC + 33.0 nC + q = -8.4 nC

q = -27.4 nC

Thus, we can conclude that charge on the third object is -27.4 nC.

7 0

3 years ago

This is one of the four planets Mercury,Venus,Earth,and Mars, whose orbits lie nearest the Sun

galben [10]

Answer:

Mercury

Explanation:

Mercury is the closest planet to the sun followed be Venus. Though Mercury is the closest Venus is actually the hotest planet

6 0

2 years ago

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A roadrunner is running along a straight desert road at a constant velocity of 25 m/s. If a certain coyote wants to capture the

andreyandreev [35.5K]

Answer:

t = 1.42 s and d = 35.5 m

Explanation:

Given that,

Velocity of a roadrunner is 25 m/s

A certain coyote wants to capture the roadrunner using a net dropped from an overpass that is 10 m high.

We need to find the time before the roadrunner is under the overpass and how far away from the overpass is the roadrunner when the coyote drops the net.

$d=ut+\dfrac{1}{2}at^2\\\\\text{Here, u = 0 and a = g}\\\\d=\dfrac{1}{2}gt^2\\\\t=\sqrt{\dfrac{2d}{g}} \\\\t=\sqrt{\dfrac{2\times 10}{9.8}} \\\\t=1.42\ s$

Let d is the distance traveled. So,

d = vt

d = 25 m/s × 1.42 s

d = 35.5 m

5 0

2 years ago

A flywheel with a diameter of 1.63 m is rotating at an angular speed of 79.9 rev/min. (a) What is the angular speed of the flywh

Studentka2010 [4]

Answer:

(a) 8.362 rad/sec

(b) 6.815 m/sec

(c) 9.446 $rad/sec^2$

(d) 396.22 revolution

Explanation:

We have given that diameter d = 1.63 m

So radius $r=\frac{d}{2}=\frac{1.63}{2}=0.815m$

Angular speed N = 79.9 rev/min

(a) We know that angular speed in radian per sec

$\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 79.9}{60}=8.362rad/sec$

(b) We know that linear speed is given by

$v=r\omega =0.815\times 8.362=6.815m/sec$

(c) We have given final angular velocity $\omega _f=675rev/min$

And $\omega _i=79.9rev/min$

Time t = 63 sec

Angular acceleration is given by $\alpha =\frac{\omega _f-\omega _i}{t}=\frac{675-79.9}{63}=9.446rad/sec^2$

(d) Change in angle is given by

$\Theta =\frac{1}{2}(\omega _i+\omega _f)t=\frac{1}{2}(675+79.9)\times 1.05=396.22rev$

7 0

2 years ago

What is the molar mass of a gas if 1.30g of the gas has a volume of 245mL at STP? ...?

padilas [110]

First, we assume this as an ideal gas so we use the equation PV=nRT. Then, we use the conditions at STP that would be 1 atm and 273.15 K. We calculate as follows:

PV= nRT
PV= mRT/MM

1 atm (.245 L) =1.30(0.08206)(273.15) / MM
MM = 118.94 g/mol <--- ANSWER

5 0

2 years ago