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3 years ago

____ is the rate of change in velocity.

Physics

2 answers:

Dimas [21]3 years ago

6 0

Answer:

Acceleration is the rate of change in velocity.

Explanation:

Velocity,v- 

"It is the change in a body's displacement or change in speed,d over the time,t."

v=s/sec,

Units: meter/second.

Acceleration,a-

"When the body has a varying velocity,v across a given time frame,t is called as the acceleration,a."

a=v/t,

Unit: meter/second².

astra-53 [7]3 years ago

4 0

The answer is Acceleration.

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To what volume will a sample of gas expand if it is heated from 50.0°c and 2.33 l to 500.0°c?

Lena [83]

The volume of the gas at $500^{\circ} \text{C}$ is $\fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}$ .

Further Explanation:

Consider the pressure of the gas to be constant.

The change in the volume of the gas when the temperature of the gas is varied by keeping the pressure of the gas at a constant value is defined by the Charles' Law.

Concept:

According to the Charles law, the volume of the gas is directly proportional to the temperature of the gas at constant pressure.

The Charles' law can be stated as:

$\fbox{\begin\\V\propto T\\\end{minispace}}$

The above expression can we written as.

$\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

Convert the temperature of the gas into kelvin.

$T=273+T^\circ\text{C}}}$

Here, T is the temperature in kelvin and $T^\circ\text{C}}}$ is the temperature in centigrade.

The initial temperature of the gas is $50^\circ\text{C}$ . The temperature of the gas in kelvin is.

$\begin{aligned}{T_1}&=273+50\\&=323\,{\text{K}}\\\end{aligned}$

The final temperature of the gas is $500^\circ\text{C}$ . The temperature in kelvin is.

$\begin{aligned}{T_2}&=273+500\\&=773\,{\text{K}}\\\end{aligned}$

Substitute the values of temperature and volume in the expression of the Charles' Law.

$\begin{aligned}{V_2}&=\frac{{{T_2}}}{{{T_1}}}{V_1}\\&=\frac{{773\,{\text{K}}}}{{323\,{\text{K}}}}\left({2.33\,{\text{L}}}\right)\\&=5.576\,{\text{L}}\\\end{aligned}$

Thus, the volume of the gas at $500^\circ\text{C}}$ will be $\fbox{\begin\ 5.576\,{\text{L}}\\\end{minispace}}$ .

Learn More:

1. Examples of wind and solar energy brainly.com/question/1062501

2. Stress developed in a wire brainly.com/question/12985068

Answer Details:

Grade: High school

Subject: Physics

Chapter: Gas law

Keywords:

Charles law, temperature, volume, initial, final, kelvin, centigrade, 50 C, 500 degree, 500 C, 50 degree, 2,33 L, gas expand, sample, heated.

7 0

3 years ago

Rebecca jogs in place for two minutes before her track meet. Rebecca weighs 100 pounds. Which of the following statements is tru

MrRissso [65]

The answer is A because it was 2 minutes and she weighs 100 lbs.... so 2(100)=200.... 200 lbs of work

7 0

3 years ago

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If you exert 250J of work onto a lever that is 5m long, what amount of force are you applying?

maks197457 [2]

From the formula of W = F·d , becuase we have the values for W and d we can find F

W = F·d
F= W/d
= 250/5
= 50 N

40 N of force was applied

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3 years ago

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Which of the following statements is true about static electricity?

natta225 [31]

Static electricity can be an alternating current.

3 0

2 years ago

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Continuous and aligned fiber-reinforced composite with cross-sectional area of 340 mm2 (0.53 in.2) is subjected to a longitudina

Alecsey [184]

(a) 23.4

The fiber-to-matrix load ratio is given by

$\frac{F_f}{F_m}=\frac{E_f V_f}{E_m V_m}$

where

$E_f = 131 GPa$ is the fiber elasticity module

$E_m = 2.4 GPa$ is the matrix elasticity module

$V_f=0.3$ is the fraction of volume of the fiber

$V_m=0.7$ is the fraction of volume of the matrix

Substituting,

$\frac{F_f}{F_m}=\frac{(131 GPa)(0.3)}{(2.4 GPa)(0.7)}=23.4$ (1)

(b) 44,594 N

The longitudinal load is

F = 46500 N

And it is sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

We can rewrite (1) as

$F_m = \frac{F_f}{23.4}$

And inserting this into (2):

$F=F_f + \frac{F_f}{23.4}$

Solving the equation, we find the actual load carried by the fiber phase:

$F=F_f (1+\frac{1}{23.4})\\F_f = \frac{F}{1+\frac{1}{23.4}}=\frac{46500 N}{1+\frac{1}{23.4}}=44,594 N$

(c) 1,906 N

Since we know that the longitudinal load is the sum of the loads carried by the fiber phase and the matrix phase:

$F=F_f + F_m$ (2)

Using

F = 46500 N

$F_f = 44594 N$

We can immediately find the actual load carried by the matrix phase:

$F_m = F-F_f = 46,500 N - 44,594 N=1,906 N$

(d) 437 MPa

The cross-sectional area of the fiber phase is

$A_f = A V_f$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_f=0.3$ , we have

$A_f = (340\cdot 10^{-6} m^2)(0.3)=102\cdot 10^{-6} m^2$

And the magnitude of the stress on the fiber phase is

$\sigma_f = \frac{F_f}{A_f}=\frac{44594 N}{102\cdot 10^{-6} m^2}=4.37\cdot 10^8 Pa = 437 MPa$

(e) 8.0 MPa

The cross-sectional area of the matrix phase is

$A_m = A V_m$

where

$A=340 mm^2=340\cdot 10^{-6}m^2$ is the total cross-sectional area

Substituting $V_m=0.7$ , we have

$A_m = (340\cdot 10^{-6} m^2)(0.7)=238\cdot 10^{-6} m^2$

And the magnitude of the stress on the matrix phase is

$\sigma_m = \frac{F_m}{A_m}=\frac{1906 N}{238\cdot 10^{-6} m^2}=8.0\cdot 10^6 Pa = 8.0 MPa$

(f) $3.34\cdot 10^{-3}$

The longitudinal modulus of elasticity is

$E = E_f V_f + E_m V_m = (131 GPa)(0.3)+(2.4 GPa)(0.7)=41.0 Gpa$

While the total stress experienced by the composite is

$\sigma = \frac{F}{A}=\frac{46500 N}{340\cdot 10^{-6}m^2}=1.37\cdot 10^8 Pa = 0.137 GPa$

So, the strain experienced by the composite is

$\epsilon=\frac{\sigma}{E}=\frac{0.137 GPa}{41.0 GPa}=3.34\cdot 10^{-3}$

3 0

3 years ago