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Vladimir79 [104]

3 years ago

9

Force F1 acts on a particle and does work W1. Force F2 acts simultaneously on the particle and does work W2. The speed of the pa

rticle does not change. Which one of the following must be true?a. W1 is zero, and W2 is positiveb. W1 = - W2c. W1 is positive, and W2 is positived. W1 is positive, and W2 is zero

1 answer:

egoroff_w [7]3 years ago

7 0

Answer:

b.) W1 = -W2

Explanation:

According to Newton's third law of motion, action and reaction are equal and opposite. For the particle to maintain a constant speed, it means that the two workdone on the particle are of eqaul magnitude but act in opposite direction. The two weights, instead of adding up annul each other and has no effect on the speed of the particle.

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Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.

Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

Explanation:

(a) Expression for change in temperature is as follows.

$|\Delta T| = |x - y|K$

= 15.1 K

= $|(x - 273.15) - (y - 273.15)|^{o}C$

= $|x - y|^{o}C$

= $15.1^{o}C$

Therefore, the magnitude of its temperature change in degrees Celsius is $15.1^{o}C$ .

(b) Change in temperature from Celsius to Fahrenheit is as follows.

F = 1.8C + 32

C = $\frac{F - 32}{1.8}$

Since, K = C + 273

or, $\Delta K = \Delta C = \frac{\Delta F}{1.8}$

$\Delta F = 1.8 \Delta K$

= 1.8 (15.1)

= $27.18^{o}F$

or, = $27.2^{o}F$

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is $27.2^{o}F$ .

7 0

2 years ago

Read 2 more answers

Positive electric charge Q is distributed uniformly throughout the volume of an insulating sphere with radius R.From the express

AlladinOne [14]

Answer:

Vb = k Q / r r <R

Vb = k q / R³ (R² - r²) r >R

Explanation:

The electic potential is defined by

ΔV = - ∫ E .ds

We calculate the potential in the line of the electric pipe, therefore the scalar product reduces the algebraic product

VB - VA = - ∫ E dr

Let's substitute every equation they give us and we find out

r> R

Va = - ∫ (k Q / r²) dr

-Va = - k Q (- 1 / r)

We evaluate with it Va = 0 for r = infinity

Vb = k Q / r r <R

We perform the calculation of the power with the expression of the electric field that they give us

Vb = - int (kQ / R3 r) dr

We integrate and evaluate from the starting point r = R to the final point r <R

Vb = ∫kq / R³ r dr

Vb = k q / R³ (R² - r²)

This is the electric field in the whole space, the places of interest are r = 0, r = R and r = infinity

8 0

3 years ago

Why are there only two elements in the first period of the periodic table?(1 point)

tester [92]

Answer:

because each row increases in atomic mass by a specific number, so anything over five is in the second row.

8 0

2 years ago

On your first trip to Planet X you happen to take along a 280 g mass, a 40-cm-long spring, a meter stick, and a stopwatch. You'r

arsen [322]

Answer:

5.31143691523 m/s²

Explanation:

m = Mass = 280 g

x = Displacement of spring = 21.7 cm

Time period

$T=\dfrac{14}{11}\\\Rightarrow T=1.27\ s$

Angular velocity is given by

$\omega=\dfrac{2\pi}{T}\\\Rightarrow \omega=\dfrac{2\pi}{1.27}\\\Rightarrow \omega=4.94739\ rad/s$

$\omega=\sqrt{\dfrac{k}{m}}\\\Rightarrow k=\omega^2m\\\Rightarrow k=4.94739^2\times 0.28\\\Rightarrow k=6.85346698739\ N/m$

From Hooke's law

$mg=kx\\\Rightarrow g=\dfrac{kx}{m}\\\Rightarrow g=\dfrac{6.85346698739\times 0.217}{0.28}\\\Rightarrow g=5.31143691523\ m/s^2$

The acceleration due to gravity on the planet is 5.31143691523 m/s²

Yes, I have been able to satisfy my curiosity.

7 0

2 years ago

James Cameron piloted a submersible craft to the bottom of the Challenger Deep, the deepest point on the ocean's floor, 11,000 m

Over [174]

Answer:

$4.1\cdot 10^8 N$

Explanation:

First of all, we need to find the pressure exerted on the sphere, which is given by:

$p=p_0 + \rho g h$

where

$p_0 =1.01\cdot 10^5 Pa$ is the atmospheric pressure

$\rho = 1000 kg/m^3$ is the water density

$g=9.8 m/s^2$ is the gravitational acceleration

$h=11,000 m$ is the depth

Substituting,

$p=1.01\cdot 10^5 Pa + (1000 kg/m^3)(9.8 m/s^2)(11,000 m)=1.08\cdot 10^8 Pa$

The radius of the sphere is r = d/2= 1.1 m/2= 0.55 m

So the total area of the sphere is

$A=4 \pi r^2 = 4 \pi (0.55 m)^2=3.8 m^2$

And so, the inward force exerted on it is

$F=pA=(1.08\cdot 10^8 Pa)(3.8 m^2)=4.1\cdot 10^8 N$

8 0

2 years ago

Read 2 more answers