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3 years ago

12

Is the term used to describe an object with enough orbital energy to escape earth completely?

1 answer:

Komok [63]3 years ago

5 0

Escape velocity describes its speed. It becomes a satellite instead of falling back to earth because as it falls under Earth’s gravity the Earth has moved so the satellite falls around the planet instead of down to the ground.

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A block sliding across a level surface has a mass of 2.5 kg and a mechanical energy of 20 joules. What is its velocity?

tia_tia [17]

Hi!

The energy of the block is 4 m/s

To calculate this, you need to use the equation for kinetic energy. The block is sliding (i.e. it's moving). If the object is sliding across a level surface, the only energy it has is kinetic energy, because there is no change in potential energy (which changes with height). So, the mechanical energy will be pure kinetic energy. The equation is the following, derived from the expression for kinetic energy:

$v= \sqrt{ \frac{2*Ke}{m}}=\sqrt{ \frac{2*20 (kg*m^{2}*s^{-2}) }{2,5kg}}=4 m/s$

Have a nice day!

8 0

3 years ago

A block M = 1.25 kg is pulled up along an inclined plane by a constant force F which is parallel to the incline as shown. The co

olga2289 [7]

F would 0.825 because the coefficient of kinetic friction between the block and the plane surface uk of 0.35

4 0

3 years ago

Convert 4.565mm to m

lutik1710 [3]

It will be 0.004565 m

4 0

4 years ago

Could somebody double check my work???

irina1246 [14]

Answer:

I think it's correct. Hope it helps.

5 0

3 years ago

A toaster using a Nichrome heating element operates on 120 V. When it is switched on at 28 ∘С, the heating element carries an in

sukhopar [10]

Answer:

The final temperature of the element = 262.67°C

The power dissipated in the heating element initially = 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A = 147.60 W

Explanation:

Our given parameters include;

A Nichrome heating element operates on 120 V.

Voltage (V) = 120V

Initial Current (I₁) = 1.36 A

Initial Temperature (T₁) = 28°C

Final Current (I₂) = 1.23 A

Final Temperature (T₂) = unknown ????

Temperature dependencies of resistance is given by:

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$ ---------------------- (1)

in which R₁ is the resistance at temperature T₁

$R_{T(2)$ is the resistance at temperature T₂

Given that V= IR

R = $\frac{V}{I}$

Therefore, the resistance at temperature 28°C is;

$R_{28}= \frac{120V}{1.36A}$

= 88.24Ω

$R_{T(2)$ = $\frac{120V}{1.23A}$

= 97.56Ω

From (1) above;

$R_{T(2)}=R_1[1+\alpha (T_2-T_1)]$

97.56 = 88.24 [ 1 + 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)]

$\frac{97.56}{88.24}= 1+(4.5*10^{-4})(T-28^0C)$

1.1056 - 1 = 4.5×10⁻⁴(°C)⁻¹(T₂-28°C)

0.1056 = 4.5×10⁻⁴(T₂-28°C)

$\frac{0.1056}{4.5*10^{-4}}= T-28^0C$

T - 28° C = 234.67

T = 234.67 + 28° C

T = 262.67 ° C

(b)

What is the power dissipated in the heating element initially and when the current reaches 1.23 A

The power dissipated in the heating element initially can be calculated as:

P = I²₁R₂₈

P = (1.36A)²(88.24Ω)

P = 163.209 W

P ≅ 163.21 W

The power dissipated in the heating element when the current reaches 1.23 A can be calculated as:

$P= I^2_2R_{T^0C$

P = (1.23)²(97.56Ω)

P = 147.598524

P ≅ 147.60 W

6 0

3 years ago