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SCORPION-xisa [38]
3 years ago
9

group of students is making model cars that will be propelled by model rocket engines. These engines provide a nearly constant t

hrust force. The cars are light—most of the weight comes from the rocket engine—and friction and drag are very small. As the engine fires, it uses fuel, so it is much lighter at the end of the run than at the start. A student ignites the engine in a car, and the car accelerates. As the fuel burns and the car continues to speed up, the magnitude of the acceleration will
Physics
1 answer:
iris [78.8K]3 years ago
7 0

Answer:

increase.

Explanation:

According to the newton’s second law of motion force is expressed as product of mass and acceleration.

F = m a

If the force acting is constant, then.

m∝ \frac{1}{a}

That is if the mass of object increases the acceleration decreases and vice versa. The above equation is used when the force acting on the body is constant.

As the thrust force from the rocket engine is constant throughout there will be a variation in the mass or acceleration.

Thus, it won't stay the same.

As the weight of the car is maximum at the start because of the fuel present in the rocket engine and minimum at the end as the fuel burns throughout the journey of the car. Weight will be minimum at the end and hence acceleration is maximum at the end.

Thus, it won't decrease.

As the acceleration is going from minimum at the start to maximum at the end, therefore it is continuously increases throughout its journey.

Thus, it will increase.  

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A drunken sailor stumbles 580 meters north, 530 meters northeast, then 480 meters northwest. What is the total displacement and
kvv77 [185]

Answer:

(a)  1294.66 m

(b) 88.44°

Explanation:

d1 = 580 m North

d2 = 530 m North east

d3 = 480 m North west

(a) Write the displacements in vector forms

\overrightarrow{d_{1}}=580\widehat{j}

\overrightarrow{d_{2}}=530\left ( Cos45\widehat{i}+Sin45\widehat{j} \right )

\overrightarrow{d_{2}}=374.77\widehat{i}+374.77\widehat{j}

\overrightarrow{d_{3}}=480\left ( - Cos45\widehat{i}+Sin45\widehat{j} \right )

\overrightarrow{d_{3}}=-339.41\widehat{i}+339.41widehat{j}

The resultant displacement is given by

\overrightarrow{d}\overrightarrow{d_{1}}+\overrightarrow{d_{2}}+\overrightarrow{d_{3}}

\overrightarrow{d}=\left ( 374.77-339.41 \right )\widehat{i}+\left ( 580+374.77+339.41 \right )\widehat{j}

\overrightarrow{d}=35.36\widehat{i}+1294.18\widehat{j}

magnitude of the displacement

d ={\sqrt{35.36^{2}+1294.18^{2}}}=1294.66 m

d = 1294.66 m

(b) Let θ be the angle from + X axis direction in counter clockwise

tan\theta =\frac{1294.18}{35.36}=36.6

θ = 88.44°

4 0
3 years ago
A copper cylinder is initially at 21.1 ∘C . At what temperature will its volume be 0.163 % larger than it is at 21.1 ∘C?
fomenos

To solve this problem we will apply the concepts related to the final volume of a body after undergoing a thermal expansion. To determine the temperature, we will use the given relationship as well as the theoretical value of the volumetric coefficient of thermal expansion of copper. This is, for example to the initial volume defined as V_1, the relation with the final volume as

V_2 = V_1 +0.163\% V_1

V_2 = V_1 +0.00163V_1

V_2 = 1.00163V_1

Initial temperature = 21.1\°C

Let T be the temperature after expanding by the formula of volume expansion

we have,

V_2 = V_1 (1+\gamma \Delta t)

Where \gamma is the volume coefficient of copper 5.1*10^{-5}/C

1.00163V_1 = V_1(1+\gamma(T-21.1\°))

1.00163 = 1+5.1*10^{-5}(T-21.1\°)

0.00163 = 0.000051T-0.0010761

T = 53.0608\°C

Therefore the temperature is 53.06°C

7 0
3 years ago
Can I improve the design of my simple machine? How?
blsea [12.9K]

Answer:

12345

Explanation:

yan na po answer ko hehehe

5 0
3 years ago
Which of the following is a false statement? Select one:
Art [367]

Answer:

The false statement is in option 'd': The center of mass of an object must lie within the object.

Explanation:

Center of mass is a theoretical point in a system of particles where the whole mass of the system is assumed to be concentrated.

Mathematically the position vector of center of mass is defined as

\overrightarrow{r}_{com}=\int \overrightarrow{r}_{i}dm

where,

\overrightarrow{r}_{i} is the position vector of the mass dm.

As we can see for homogenous symmetrical objects such as a sphere,cube,disc the center of mass is located at the centroid of the shapes itself but in many shapes it is located outside the body also.

Examples of shapes in which center of mass is located outside the body:

1) Horseshoe shaped body.

2) A thin ring.

In many cases we can make shapes of bodies whose center of mass lies outside the body.

6 0
3 years ago
Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determi
Talja [164]
The solution to the problem is as follows:

 <span>Average = 80 
So Sum = 80 * 5 = 400 
Mode = 88, so two results are 88 (if three results were 88, then the median would be 88). 
Three results are 81, 88, and 88. 
That leaves 143. We could still have one 81 score, so that leaves the lowest score as 62. 

Greg is in a car at the top of a roller-coaster ride. The distance, d, of the car from the ground as the car descends is determined by the equation d = 144 - 16t2, where t is the number of seconds it takes the car to travel down to each point on the ride. How many seconds will it take Greg to reach the ground? 
d = 144 - 16t2 
0 = 144 - 16t2 
16t^2=144 
t^2=9 
t=3</span>
3 0
3 years ago
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