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Ugo [173]
10 months ago
13

A motorcycle moving at 18 m/s decelerates at a rate of 3.6 m/s². What is the car's

Physics
1 answer:
Gekata [30.6K]10 months ago
5 0

Answer:

V = 3.6 m/s

Explanation:

Given:

V₀ = 18 m/s

a = - 3,6 m/s²  (The motorcycle is slowing down!)

t = 4 s

____________

V - ?

Motorcycle speed:

V = V₀ + a·t

V = 18 + (-3.6)·4 = 3.6 m/s

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Potential diffetence​
Stells [14]

Answer:

6v

Explanation:

V=IR

V= 2* 3

V= 6 volts

7 0
3 years ago
What is net force? A. The fundamental forces B. The unified force C. The gravitational force D. The vector sum of the forces
avanturin [10]
It is the vector sum of the forces
8 0
3 years ago
Read 2 more answers
A 50 g copper calorimeter contains 250 g of water at 20 C. How much steam be condensed into the water to make the final temperat
Nostrana [21]

Answer:

Approximately 13\; \rm g of steam at 100\; \rm ^\circ C (assuming that the boiling point of water in this experiment is 100\; \rm ^\circ C\!.)

Explanation:

Latent heat of condensation/evaporation of water: 2260\; \rm J \cdot g^{-1}.

Both mass values in this question are given in grams. Hence, convert the specific heat values from this question to \rm J \cdot g^{-1}.

Specific heat of water: 4.2\; \rm J \cdot g^{-1}\cdot \rm K^{-1}.

Specific heat of copper: 0.39\; \rm J \cdot g^{-1}\cdot K^{-1}.

The temperature of this calorimeter and the 250\; \rm g of water that it initially contains increased from 20\; \rm ^\circ C to 50\; \rm ^\circ C. Calculate the amount of energy that would be absorbed:

\begin{aligned}& Q(\text{copper}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 0.39\; \rm J \cdot g^{-1}\cdot K^{-1} \times 50\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 585\; \rm J  \end{aligned}.

\begin{aligned}& Q(\text{cool water}) \\ =\;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; \rm J \cdot g^{-1}\cdot K^{-1} \times 250\; \rm g \times (50\;{\rm ^\circ C} - 20\;{\rm ^\circ C}) \\ =\; & 31500\; \rm J  \end{aligned}.

Hence, it would take an extra 585\; \rm J + 31500\; \rm J = 32085\; \rm J of energy to increase the temperature of the calorimeter and the 250\; \rm g of water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

Assume that it would take x grams of steam at 100\; \rm ^\circ C ensure that the equilibrium temperature of the system is 50\; \rm ^\circ C.

In other words, x\; \rm g of steam at 100\; \rm ^\circ C would need to release 32085\; \rm J as it condenses (releases latent heat) and cools down to 50\; \rm ^\circ C.

Latent heat of condensation from x\; \rm g of steam: 2260\; {\rm J \cdot g^{-1}} \times (x\; {\rm g}) = (2260\, x)\; \rm J.

Energy released when that x\; {\rm g} of water from the steam cools down from 100\; \rm ^\circ C to 50\; \rm ^\circ C:

\begin{aligned}Q = \;& c \cdot m \cdot \Delta t \\ =\;& 4.2\; {\rm J \cdot g^{-1}\cdot K^{-1}} \times (x\; \rm g) \times (100\;{\rm ^\circ C} - 50\;{\rm ^\circ C}) \\ =\; & (210\, x)\; \rm J  \end{aligned}.

These two parts of energy should add up to 32085\; \rm J. That would be exactly what it would take to raise the temperature of the calorimeter and the water that it initially contains from 20\; \rm ^\circ C to 50\; \rm ^\circ C.

(2260\, x)\; {\rm J} + (210\, x)\; {\rm J} = 32085\; \rm J.

Solve for x:

x \approx 13.

Hence, it would take approximately 13\; \rm g of steam at 100\; \rm ^\circ C for the equilibrium temperature of the system to be 50\; \rm ^\circ C.

4 0
2 years ago
Please help me pwease
Alisiya [41]

Answer:

Iron

Explanation:  It is by mass the most common element on Earth, right in front of oxygen, forming much of Earth's outer and inner core.

6 0
2 years ago
What is the measure of the change in velocity during the period of time?
Rashid [163]
That measure is known as "acceleration"
a = ΔV / t

Hope this helps!
3 0
3 years ago
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