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Kobotan [32]
3 years ago
12

On Earth, the gravitational field strength is 10 N/kg. Calculate E for a 4 kg bowling ball that is being

Physics
1 answer:
Zarrin [17]3 years ago
3 0

Answer:

People are exposed to sources of radiation in all aspects of everyday life. Radioactive sources can be very useful but need handling carefully to ensure safety.

Explanation:

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D. 'g' vanishes at centre of
guapka [62]

Answer:

a. 0.8 cm

Explanation:

The distance of the object from the lens, u = 1 cm

The magnification of the lens, m = 5

The focus of a lens formula is given as follows;

f = \dfrac{1}{\dfrac{1}{v} + \dfrac{1}{v}  }

The magnification of the lens, m = -v/u

Where;

v = The distance of the image from the lens

Therefore, we have;

v = m × u

∴ v = 5 × 1 cm = 5 cm (on the other side of the lens)

From which we get;

f = \dfrac{1}{\dfrac{1}{1} + \dfrac{1}{5}  } = \dfrac{5}{6}  \approx 0.8

The focal length ≈ 0.8 cm

6 0
3 years ago
Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by th
Andrej [43]

Answer:

The charge carried by the droplet is 1.330\times10^{-19}\ C

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude 5.92\times10^4\ N/C points straight down and if the mass of the droplet is 2.93\times10^{-15} kg

We need to calculate the acceleration

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2

a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}

a=2.688\ m/s^2

We need to calculate the charge carried by the droplet

Using formula of electric filed

E=\dfrac{F}{q}

q=\dfrac{ma}{E}

Put the value into the formula

q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}

q=1.330\times10^{-19}\ C

Hence, The charge carried by the droplet is 1.330\times10^{-19}\ C

8 0
3 years ago
If this were a theoretical frictionless plane, what would be the mechanical advantage?
alex41 [277]
Mechanical advantage (MA = slope/height)
here length of slope = 9Hheight = H                                                 so mechanical advantage = ---- 9H/H= 9                                                  
3 0
3 years ago
A fixed 15.3-cm-diameter wire coil is perpendicular to a magnetic field 0.77 T pointing up. In 0.20 s , the field is changed to
maksim [4K]

Answer:

Induced emf will be 0.468 volt

Explanation:

We have given diameter of wire d = 15.3 cm

So radius r=\frac{d}{2}=\frac{15.3}{2}=7.65cm

So area A=\pi r^2=3.14\times (7.65)^2=183.76\times 10^{-4}m^2

Change in magnetic field dB = 0.26 - 0.77 = -0.51 T

Time for change in magnetic field dt = 0.26 sec

We know that emf is given by e=\frac{-d\Phi }{dt}=-A\frac{dB}{dt}=-183.76\times 10^{-4}\times \frac{-0.51}{0.2}=0.0468volt

5 0
3 years ago
Help quick (edge 2021)
kow [346]

Answer:

a

Explanation:

8 0
3 years ago
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