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pickupchik [31]
4 years ago
15

A car, moving along a straight stretch of highway,

Physics
1 answer:
yarga [219]4 years ago
5 0
<span>We know that an object in moving with acceleration follow the rule according that the distance covered will be : d = Vi*t + 1/2*a*t^2 where d is distance, Vi is initial speed, and a is acceleration Then after 1 km which is 1000 metres we have: 1000 = Vi *71.2 + 1/2*0.0499*(71.2)^2 Vi = (1000-1/2*0.0499*(71.2)^2)/71.2 = 1000/71.2 - 1/2*0.0499*71.2 = 12.27 m/s Then the car was going at 12.27 m/s when started to accelerate.</span>
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Large radio telescopes, like the one in Arecibo, Puerto Rico, can detect extremely weak signals. Suppose one radio telescope is
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Explanation:

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3 years ago
Which metals have similar chemical properties? Check all that apply.
Alika [10]

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3 0
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3.12 ** To illustrate the use of a multistage rocket consider the following: (a) A certain rocket carries 60% of its initial mas
madreJ [45]

Answer:

0.916v_{ex}

Explanation:

A multistage rock is also known as a step rocket and it is a form of vehicle that can use more than a rocket stage. The rocket stage typically contains propellant as well as its engine. The final speed of the rocket can be estimated using the equation below:

v = v_{ex} ln(\frac{m_{o}}{0.4m_{o}}) = v_{ex}*ln(2.5) = v_{ex}*0.916 = 0.916v_{ex}

Therefore, the maximum speed of the rocket is 0.916v_{ex}

8 0
4 years ago
Mass of a body is 210g and its density is 7.981g/cm^3 what will be its volume with regard to significant figures
aleksandrvk [35]

Density is mass per unit volume,

\rho=\dfrac mV

so to get the volume, divide the mass by the density,

V=\dfrac m\rho

So the volume is

\dfrac{210\,\mathrm g}{7.981\frac{\rm g}{\mathrm{cm}^3}}\approx26.312\,\mathrm{cm}^3\approx\boxed{26\,\mathrm{cm}^3}

3 0
3 years ago
hot water is added to three times its mass of water at 10 degree celsius and the resulting temperature is 20 degrees Celsius Wha
olchik [2.2K]

Answer:

The initial temperature of the hot water is 50\; \rm ^{\circ} C (assuming that no heat was lost to the surroundings.)

Explanation:

Let m denote the mass of the hot water.

The question states that the mass of the water at 10\; \rm ^\circ C is three times the mass of the hot water. If the mass of the hot water is m, the mass of the cold water would be 3\, m.

Let c denote the specific heat capacity of water. Let m denote the mass of some water. The energy required to change the temperature of that much water by \Delta T (without state change) would be:

Q = c \cdot m \cdot \Delta T.

The temperature change for the cold water was:

\Delta T_1 = 20\; \rm ^{\circ} C - 10\; \rm ^{\circ} C = 10\; \rm K.

Energy required to raise the temperature of water with mass 3\, m from 10\; \rm ^{\circ} C to 20\; \rm ^{\circ} C:

Q_1 = c \cdot (3\, m) \cdot (10\; \rm K).

On the other hand, if the initial temperature of the hot water is t\; \rm ^{\circ} C (where t > 20,) the temperature change would be:

\Delta T_2 = t\; {\rm ^{\circ} C} - 20\; {\rm ^{\circ} C} = (t - 20)\; {\rm K}.

Calculate the energy change involved:

Q_2 = c \cdot m \cdot ((t - 20)\; \rm K).

If no energy was lost to the surroundings, Q_1 should be equal to Q_2. That is:

c \cdot (3\, m) \cdot (10\; {\rm K}) = c\cdot m \cdot ((t - 20)\; {\rm K}).

Simplify and solve for t:

t - 20 = 30.

t = 50.

Therefore, the initial temperature of the hot water would be 50\; {\rm ^\circ C}.

6 0
3 years ago
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