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3 years ago

The rate of change of an object's velocity is called _______.

Physics

2 answers:

tamaranim1 [39]3 years ago

4 0

Answer:

The rate of change of an object's velocity is called acceleration.

aleksklad [387]3 years ago

3 0

It is called acceleration

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A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 563 K, and r

Schach [20]

Answer:

1.85 J/K

Explanation:

The computation of total change in entropy is shown below:-

Change in Entropy = Sum Q ÷ T

= $\frac{-heat\ entering\ the\ reservoir}{Reservoir\ 1\ Temperature} + \frac{Conduction\ of\ heat}{Reservoir\ 2\ Temperature}$

$= \frac{-1760}{563} + \frac{1760}{354}$

= -3.12 + 4.97

= 1.85 J/K

Therefore for computing the total change in entropy we simply applied the above formula.

As we can see that there is heat entering the reservoir so it will be negative while cold reservoir will be positive else the process would be impossible.

8 0

4 years ago

A 6,000N is applied to a formula one car that weighs 500kg. What is the car's acceleration?

Vesna [10]

Answer:

Explanation:

The acceleration of an object given it's mass and the force acting on it can be found by using the formula

$a = \frac{f}{m} \\$

f is the force

m is the mass

From the question

f = 6000 N

m = 500 kg

We have

$a = \frac{6000}{500} = \frac{60}{5} = 12 \\$

We have the final answer as

Hope this helps you

8 0

3 years ago

A ball with a volume of 3 cm3 and a density of 9.0 g/cm3 increases its velocity from 2 m/s to 6 m/s over a 12

V125BC [204]

Answer:

M = ρ V = 9 gm/cm^ 3 * cm^3 = 27 gm

a = (V2 - V1) / t = (6 - 2) m/s / 12 s = 1/3 m/s^2 the acceleration

F = M a = 27 gm * 1/3 m/s^2 = 9 dynes net force applied

4 0

2 years ago

A 60.0-kg skier with an initial speed of 12.0 m/s coasts up a 2.50-mhigh rise and angle is 35 degrees.Find her final speed at th

Shalnov [3]

Wt. = Fg = m*g = 60kg * 9.8N/kg=588 N.=
Wt. of skier. 

Fp=588*sin35 = 337 N.=Force parallel to 
incline. 
Fv = 588*cos35 = 482 N. = Force perpendicular to incline. 

Fk = u*Fv = 0.08 * 482 = 38.5 N. = Force 
of kinetic friction. 
d =h/sinA = 2.5/sin35 = 4.36 m. 

Ek + Ep = Ekmax - Fk*d 
Ek = Ekmax-Ep-Fk*d 
Ek=0.5*60*12^2-588*2.5-38.5*4.36=2682 J. 
Ek = 0.5m*V^2 = 2682 J. 
30*V^2 = 2682 
V^2 = 89.4 
V = 9.5 m/s = Final velocity.

4 0

4 years ago

Read 2 more answers

A roller coaster car is traveling at a constant 3 m/s when it reaches a downward slope. On the slope, the car accelerates at a c

kirill [66]

Answer:

Velocity of the car at the bottom of the slope: approximately $20.3\; \rm m \cdot s^{-2}$ .

It would take approximately $3.9\; \rm s$ for the car to travel from the top of the slope to the bottom.

Explanation:

The time of the travel needs to be found. Hence, make use of the SUVAT equation that does not include time.

Let $v$ denote the final velocity of the car.
Let $u$ denote the initial velocity of the car.
Let $a$ denote the acceleration of the car.
Let $x$ denote the distance that this car travelled.

$v^2 - u^2 = 2\, a\cdot x$ .

Given:

$u = 3\; \rm m \cdot s^{-1}$ .
$a = 4.5\; \rm m \cdot s^{-2}$ .
$x = 45\; \rm m$ .

Rearrange the equation $v^2 - u^2 = 2\, a\cdot x$ and solve for $v$ :

$\begin{aligned}v &= \sqrt{2\, a \cdot x + u^2} \\ &= \sqrt{2 \times 4.5\; \rm m \cdot s^{-2} \times 45\; \rm m + \left(3\; \rm m \cdot s^{-1}\right)^{2}} \\ &\approx 20.3\; \rm m \cdot s^{-1}\end{aligned}$ .

Calculate the time required for reaching this speed from $u = 3\; \rm m \cdot s^{-1}$ at $a = 4.5\; \rm m \cdot s^{-2}$ :

$\begin{aligned}t &= \frac{v - u}{a} \\ &\approx \frac{20.3\; \rm m \cdot s^{-1} - 3\; \rm m \cdot s^{-1}}{4.5\; \rm m \cdot s^{-2}} \approx 3.9\; \rm m \cdot s^{-1}\end{aligned}$ .

3 0

3 years ago