All categories

2 years ago

PLEASE HELP ME!!!!!!!!

Physics

2 answers:

Ray Of Light [21]2 years ago

5 0

Answer:

I believe C

Explanation:

i never learned this sorry

DerKrebs [107]2 years ago

3 0

The answer is C) if I’m correct

You might be interested in

Waves with higher amplitude transfer ____energy.

natali 33 [55]

Answer:

- Waves with higher amplitude transfer HIGHER energy.

- Waves with higher frequency transfer HIGHER energy.

3 0

2 years ago

A 210 g block is dropped onto a relaxed vertical spring that has a spring constant of k = 2.0 N/cm. The block becomes attached t

Yuliya22 [10]

Answer:

a) $W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

b) $W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

c) $V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

d) $d_1 =0.183m$ or 18.3 cm

Explanation:

For this case we have the following system with the forces on the figure attached.

We know that the spring compresses a total distance of x=0.10 m

Part a

The gravitational force is defined as mg so on this case the work donde by the gravity is:

$W_{g}=mdx = 0.21 kg *9.8\frac{m}{s^2} 0.10m=0.2058 J$

Part b

For this case first we can convert the spring constant to N/m like this:

$2 \frac{N}{cm} \frac{100cm}{1m}=200 \frac{N}{m}$

And the work donde by the spring on this case is given by:

$W_{spring}= -\frac{1}{2} Kx^2 =-\frac{1}{2} 200 N/m (0.1m)^2=-1 J$

Part c

We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

$W_{g} +W_{spring} = K_{f} -K_{i}=0- \frac{1}{2} m v^2_i$

And if we solve for the initial velocity we got:

$V_i =\sqrt{2 \frac{W_g + W_{spring}}{0.21 kg}}}=\sqrt{2 \frac{(1-0.2058)}{0.21 kg}}}=2.75m/s$

Part d

Let d1 represent the new maximum distance, in order to find it we know that :

$-1/2mV^2_i = W_g + W_{spring}$

And replacing we got:

$-1/2mV^2_i =mg d_1 -1/2 k d^2_1$

And we can put the terms like this:

$\frac{1}{2} k d^2_1 -mg d_1 -1/2 m V^2_i =0$

If we multiply all the equation by 2 we got:

$k d^2_1 -2 mg d_1 -m V^2_i =0$

Now we can replace the values and we got:

$200N/m d^2_1 -0.21kg(9.8m/s^2)d_1 -0.21 kg(5.50 m/s)^2) =0$

$200 d^2_1 -2.058 d_1 -6.3525=0$

And solving the quadratic equation we got that the solution for $d_1 =0.183m$ or 18.3 cm because the negative solution not make sense.

5 0

2 years ago

A certain ideal gas has molar heat capacity at constant volume CV. A sample of this gas initially occupies a volume V0 at pressu

ANTONII [103]

Answer:

Explanation:

The processes are described on the image attached below. The isobaric process consists of an horizontal line, the adiabatic expansion is described by a polytropic curve:

$P_{2} \cdot V_{2}^{\gamma} = P_{3} \cdot V_{3}^{\gamma}$

Where:

$\gamma = \frac{c_{p}}{c_{v}}$

$\gamma = 1 + \frac{R}{c_{v}}$

Final pressure is:

$P_{3} = P_{2}\cdot \left(\frac{V_{2}}{V_{3}} \right)^{\gamma}$

$P_{3} = P_{o}\cdot \left(\frac{1}{2}\right)^{\gamma}$

$P_{3} = \frac{P_{o}}{2^{\gamma}}$

8 0

2 years ago

How many protons would the element with the atomic number 10 contain?

galben [10]

The atomic number is the same as the proton number so the answer would be D) 10

3 0

2 years ago

Imagine a large block that sinks in a tub of water. You hold a much smaller block of the exact same material in your hand, above

goldenfox [79]

It will sink because mass does not affect the physical properties of the object.

3 0

2 years ago