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Bogdan [553]
2 years ago
8

In this lab you will use a cart and track to explore various aspects of motion. You will measure and record the time it takes th

e cart to travel certain distances, and then complete some calculations. In the space below, write a scientific question that you will answer by doing this experiment.
Physics
2 answers:
andreyandreev [35.5K]2 years ago
8 0

Answer:

How does an object's position and velocity change as the object accelerates?

Explanation:

Keith_Richards [23]2 years ago
3 0

It is given that by using track and cart we can record the time and the distance travelled and also the speed of the cart can be recorded. With all this data we can solve questions on the laws of motion.  

Like using the first law of motion we can determine the force of gravity acting on the cart that has moved a certain distance and the velocity or the speed of card has already been registered and since time is known putting the values in formula would help us calculate the gravitational pull acting on cart.

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The position of a particle moving along the x axis may be determined from the expression x(t) = btu + ctv, where x will be in me
KIM [24]

As per given equation we have

x = bt^u + ct^v

now as per the dimensional analysis we can say that dimension of right side of equation must be equal to left side of the equation

now as per left side of equation its dimension is same as length or meter

now we can say it should be meter on right side also

bt^u = M^0L^1T^0

b*T^8 = M^0L^1T^0

b = M^0L^1T^{-8}

similarly for other term we have

ct^v = M^0L^1T^0

c*T^7 = M^0L^1T^0

c = M^0L^1T^{-7}

<em>so above are the dimensions of b and c</em>

8 0
1 year ago
A gas is compressed from 600 cm3 to 200cm3 at a constant pressure of 400 kpa. at the same time, 100 j of heat energy is transfer
Mekhanik [1.2K]
The initial volume of the gas is
V_i = 600 cm^3
while its final volume is
V_f = 200 cm^3
so its variation of volume is
\Delta V = V_f - V-i = 200 cm^3 - 600 cm^3 = -400 cm^3 = -400 \cdot 10^{-6} m^3

The pressure is constant, and it is
p=400 kPa = 400 \cdot 10^3 Pa

Therefore the work done by the gas is
W=p\Delta V = (400 \cdot 10^3 Pa)(-400 \cdot 10^{-6} m^3)=-160 J
where the negative sign means the work is done by the surrounding on the gas.

The heat energy given to the gas is
Q=+100 J

And the change in internal energy of the gas can be found by using the first law of thermodynamics:
\Delta U = Q-W = 100 J - (-160 J)=+260 J
where the positive sign means the internal energy of the gas has increased.
7 0
2 years ago
A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th
Tju [1.3M]

Answer:

a = \frac{2}{3}g

T = \frac{mg}{3}

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

mg - T = ma

also by torque equation we can say

TR = I\alpha

TR = \frac{1}{2}mR^2(\frac{a}{R})

T = \frac{1}{2}ma

Now we have

mg - \frac{1}{2}ma = ma

mg = \frac{3}{2}ma

a = \frac{2}{3}g

Also from above equation the tension force in the string is

T = \frac{1}{2}ma

T = \frac{mg}{3}

7 0
2 years ago
A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove
True [87]

Answer:t=5.07 s

Explanation:

Given

height of Building h=18 m

mass of Paint can m_1=4 kg

mass of second can m_2=3 kg

let T be the Tension in the rope

For  4 kg can

4g-T=4a

T=4(g-a)----1

For 3 kg can

T-3g=3a

T=3(g+a)----2

From 1 and 2

4(g-a)=3(g+a)

g=7a

a=\frac{g}{7}

So time taken to cover 18 m is

h=ut+\frac{at^2}{2}

18=0+\frac{g\cdot t^2}{7\times 2}

t^2=\frac{18\times 2\times 7}{g}

t=5.07 s

5 0
2 years ago
Let’s now multiply two numbers in scientific notation using Google.
olga_2 [115]
We are going to rewrite each number:
 (4.48E-8) = 0.0000000448
 (5.2E-4) = 0.00052
 We observe that when multiplying, the exponent will be on the order of 10 ^ -11
 Doing the multiplication we have:
 (4.48E-8) * (5.2E-4) = 2.3296E-11
 Rewriting:
 (4.48E-8) * (5.2E-4) = 2.33E-11
 Answer:
 
2.33E-11
5 0
2 years ago
Read 2 more answers
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