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frozen [14]
3 years ago
14

If a cart goes around a turn at 20 km/h ,what remains constant

Physics
1 answer:
Vitek1552 [10]3 years ago
7 0

Answer: 4.speed

Explanation:

In this case, we know that the cart remains at a constant 20km/h.

Now, one could say that "the velocity remains constant, because it always is 20km/h"

But remember that velocity is a vector, so this has a direction, and if the cart is going around a turn, then the direction of motion is changing, which tell us that there is acceleration.

But the module of the velocity, the speed, remains constant at 20km/h.

Then the correct option is 4, speed.

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Which graph best represents the relationship between the conductivity and the temperature of semiconductors ?​
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Answer: I think the answer is B

Explanation:

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how does spatial pattern of heights illustrate the relationship between temperature density and the rate of vertical pressure ch
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The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

Generally, the relationship between temperature, density and rate of vertical pressure is given as;

\rho = \frac{PM}{RT}

\frac{dP}{dz} = -\rho g\\\\

where;

  • <em>ρ is density</em>
  • <em>T is temperature</em>
  • <em>dP is rate of change of vertical  pressure</em>

Thus, from the formula above, we can conclude the following relationship between temperature, density and the rate of vertical pressure change in spatial pattern of heights.

The rate of change of vertical pressure is directly proportional to density and also directly proportional to temperature.

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3 years ago
The coefficient of linear expansion of copper is 17 x 10^-6 K-1. A block of copper 30 cm wide, 45 cm long, and 10 cm thick is he
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Answer:

The change in volume is 6.885\times 10^{- 5}\

Solution:

As per the question:

Coefficient of linear expansion of Copper, \alpha = 17\times 10^{- 6}\ K^{- 1}

Initial Temperature, T = 0^{\circ} = 273 K

Final Temperature, T' = 100^{\circ} = 273 + 100 = 373 K

Now,

Initial Volume of the block, V = 30\times 45\times 10\times 10^{- 6}\ m^{3} = 0.0135\ m^{3}

V' = V(1 + \gamma \Delta T)

\gamma = 3\alpha

V' = V(1 + 3\alpha \Delta T)

where

V' = Final volume

V' - V= 0.0135\times 17\times 10^{- 6} \times (T' - T))

\Delta V= 0.0135\times 3\times 17\times 10^{- 6} \times (373 - 273)) = 6.885\times 10^{- 5}\

7 0
3 years ago
Calculate how much work is required to launch a spacecraft of mass mm from the surface of the earth (mass mEmE, radius RERE) and
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Answer:

Work done = (1/2)[(Gmm_e)/(R_e)]

Explanation:

I've attached the explanations below.

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A motorcycle moving 18.8 m/s has<br> 57800 J of KE. What is its mass?
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Answer:

m = 327.07 kg

Explanation:

Given that,

Kinetic energy of a motorcycle, E = 57800 J

Velocity of the motorcycle, v = 18.8 m/s

We need to find the mass of the motorcycle. The kinetic energy of an object is given by :

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So, the mass of the motorcycle is 327.07 kg.

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3 years ago
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