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natita [175]
3 years ago
9

Which is greater, the gravitational force between earth and the moon, or the force between earth and the sun?

Physics
1 answer:
Karo-lina-s [1.5K]3 years ago
8 0
Earth and the Sun
GF = 3.647x10^22 N
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If chilled coke and hot tea are kept together tea cools down but coke gets warm .why?​
Harlamova29_29 [7]

When hot tea is mixed to chilled coke, tea loses heat and coke gains heat. Thus, tea cools down but coke gets heated. Because it is liquid and liquid does not totally cool down to the ambient temperature, it and the iced drink will eventually reach the same temperature.

6 0
2 years ago
A curve of radius 53.1 m is banked so that a car of mass 2.9 Mg traveling with uniform speed 67 km/hr can round the curve withou
prisoha [69]

Answer:

33.65°

Explanation:

radius, r = 53.1 m

m = 2.9 Mg = 2.9 x 10^6 g = 2900 kg

v = 67 km/h

convert km/h into m/s

v = 18.61 m/s

Let the angle of banking of road is θ, without friction

tan\theta =\frac{v^{2}}{rg}

tan\theta =\frac{18.61^{2}}{53.1\times 9.8}

tan  θ = 0.6655

θ = 33.65°

Thus, the angle of banking of road is 33.65°.

6 0
3 years ago
2 76 Consider a household that uses 14,000 kWh of elec- (tricity per year and 3400 L of fuel oil during a heating sea- son. The
NNADVOKAT [17]

Answer:

The total amount of CO₂ produced will be  = 20680 kg/year

The reduction in the amount of CO₂ emissions by that household per year = 3102 kg/year

Explanation:

Given:

Power used by household = 14000 kWh

Fuel oil used = 3400 L

CO₂ produced of fuel oil = 3.2 kg/L

CO₂ produced of electricity = 0.70 kg/kWh

Now, the total amount of CO₂ produced will be = (14000 kWh × 0.70 kg/kWh) + (3400 L × 3.2 kg/L)

⇒ The total amount of CO₂ produced will be = 9800 + 10880 = 20680 kg/year

Now,

if the usage of electricity and fuel oil is reduced by 15%, the reduction in the amount of the CO₂ emission will be = 0.15 × 20680 kg/year = 3102 kg/year

6 0
3 years ago
Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accele
lyudmila [28]

Answer:

Part a)

v = 3.16 \times 10^7 m/s

Part b)

r = 0.166 m

Explanation:

Part a)

As we know that the energy of the Hydride ion is given as

E = 5 MeV

here we have

\frac{1}{2}mv^2 = 5\times 10^6(1.6 \times 10^{-19})

also we know that

m = 1.6 \times 10^{-27} kg

now we have

v = \sqrt{\frac{2 \times 5 \times 10^6(1.6 \times 10^{-19}}{1.6\times 10^{-27}}

v = 3.16 \times 10^7 m/s

Part b)

As we know that magnetic force on the charge is centripetal force

so we have

qvB = \frac{mv^2}{r}

so we have

r = \frac{mv}{qB}

so we have

r = \frac{1.6 \times 10^{-27}(3.16 \times 10^7)}{(1.6 \times 10^{-19}) 1.9}

r = 0.166 m

4 0
3 years ago
A 4 kg block is moving at 12 m/s on a horizontal frictionless surface. a constant force is applied such that the block slows wit
weeeeeb [17]

Answer:

Is there a picture?

Explanation:

8 0
3 years ago
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