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3 years ago

1.

Physics

2 answers:

charle [14.2K]3 years ago

8 0

Answer:

i love devon bostick

Explanation:

cause....... i do

Semmy [17]3 years ago

7 0

Answer: It was going 4 m/s

Explanation:

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A 0.750 kg block is attached to a spring with spring constant 13.0 N/m . While the block is sitting at rest, a student hits it w

trapecia [35]

To solve this problem we will apply the concepts related to energy conservation. From this conservation we will find the magnitude of the amplitude. Later for the second part, we will need to find the period, from which it will be possible to obtain the speed of the body.

A) Conservation of Energy,

$KE = PE$

$\frac{1}{2} mv ^2 = \frac{1}{2} k A^2$

Here,

m = Mass

v = Velocity

k = Spring constant

A = Amplitude

Rearranging to find the Amplitude we have,

$A = \sqrt{\frac{mv^2}{k}}$

Replacing,

$A = \sqrt{\frac{(0.750)(31*10^{-2})^2}{13}}$

$A = 0.0744m$

(B) For this part we will begin by applying the concept of Period, this in order to find the speed defined in the mass-spring systems.

The Period is defined as

$T = 2\pi \sqrt{\frac{m}{k}}$

Replacing,

$T = 2\pi \sqrt{\frac{0.750}{13}}$

$T= 1.509s$

Now the velocity is described as,

$v = \frac{2\pi}{T} * \sqrt{A^2-x^2}$

$v = \frac{2\pi}{T} * \sqrt{A^2-0.75A^2}$

We have all the values, then replacing,

$v = \frac{2\pi}{1.509}\sqrt{(0.0744)^2-(0.750(0.0744))^2}$

$v = 0.2049m/s$

7 0

3 years ago

What are some examples of aerobic exercise?

Firdavs [7]

Answer:

swimming, cycling, and jogging

Explanation:

4 0

3 years ago

A container with volume 1.64 L is initially evacuated. Then it is filled with 0.226 g of N2N

vaieri [72.5K]

Answer:

0.015 atm

Explanation:

The pressure of the gas can be calculated using Ideal Gas Law:

$p = \frac{nRT}{V}$

Where:

n: is the number of moles of the gas

R: is the gas constant = 0.082 L*atm/(K*mol)

V: is the volume of the container = 1.64 L

T: is the temperature

We need to find the number of moles and the temperature. The number of moles is:

$n = \frac{m}{M}$

Where:

M: is the molar mass of the N₂ = 14.007 g/mol*2 = 28.014 g/mol

m: is the mass of the gas = 0.226 g

$n = \frac{0.226 g}{28.014 g/mol} = 8.07 \cdot 10^{-3} moles$

Now, the temperature can be found using the following equation:

$v_{rms} = \sqrt{\frac{3RT}{M}}$

Where:

R: is the gas constant = 0.082 L*atm/K*mol = 8.314 J/K*mol

$v_{rms}$ : is the root-mean-square speed of the gas = 182 m/s

By solving the above equation for T, we have:

$T = \frac{v_{rms}^{2}*M}{3R} = \frac{(182 m/s)^{2}*28.014 \cdot 10^{-3} Kg/mol}{3*8.314 J K^{-1}mol^{-1}} = 37.20 K$

Finally, we can find the pressure of the gas:

$p = \frac{nRT}{V} = \frac{8.07 \cdot 10^{-3} mol*0.082 L*atm* K^{-1}*mol^{-1}*37.20 K}{1.64 L} = 0.015 atm$

Therefore, the pressure of the gas is 0.015 atm.

I hope it helps you!

8 0

3 years ago

Un automóvil acelera de 0 a 140 km/h en 9.8 segundos, determina su aceleración

guajiro [1.7K]

Answer:

The acceleration is 14.28 km/h^2

Explanation:

Step one:

Given data

initial speed u= 0 km/h

final speed v= 140km/h

time t= 9.8 seconds

Required

The acceleration of the car

Step two:

From a= v-u/t

substitute

a= 140-0/9.8

a=140/9.8

a=14.28 km/h^2

6 0

3 years ago

After an initial race George determines that his car loses 35 percent of its acceleration due to air resistance travelling at 38

ddd [48]

Answer:

64.2 m/s

Explanation:

We are given that

Speed ,v=38 m/s

We have to find the maximum speed when his car reach on flat ground.

Using dimensional analysis

$F_{res}\propto v^2$

If 35% acceleration reduced by F(res) at 38 m/s

Then, 100% acceleration can be reduced by F(res) at v' m/s

$\frac{F_1}{F_2}=\frac{v^2}{v'^2}$

$v'^2=\frac{F_2}{F_1}v^2$

$v'=v\sqrt{\frac{F_2}{F_1}}$

Substitute the values

$v'=38\times \sqrt{\frac{100}{35}}$

$v'=64.2 m/s$

Hence, the maximum speed when his car can reach on flat ground=64.2 m/s

3 0

3 years ago