The slope of a distance-versus-time graph shows an object's

State the law of universal gravitation, and use examples to explain how changes in mass and changes in distance affect gravitati

juin [17]

F(g)= Gm1m2/ r^2 If mass is increased, so will the force of gravity because it is in direct relationship with the gravitational force, but if distance is increased, the force of gravity will decrease because it is indirectly related ( since it is on the bottom of the equation)

6 0

3 years ago

(LC)Light fixtures and placement that create shadows on the set, that obscure or completely hide action in certain areas of the

Sedaia [141]

The correct answer is true.

It is true that light fixtures and placement that create shadows on the set, that obscure or completely hide action in certain areas of the set, or that change as the main character’s emotional state changes are all ways that lighting can be used to heighten the drama and suspense in dramatic films.

Lighting plays an important role in film making because it can create scenes that enhance the de drama of the moment or the right mood that the director wants to share. Lighting in the film is an art because the basic principle is that the scene needs to look natural. From that principle, filmmakers and light specialist cand create many kinds of dramatic or jubilation moments if they know how to apply light principles to each scene.

5 0

3 years ago

In a standard tensile test, a steel rod of 7 8-in. diameter is subjected to a tension force of 17 kips. Knowing that ν = 0.30 an

natali 33 [55]

Answer:

(a) Elongation of the rod==5.61×10⁻⁹m

(b) Change in diameter=1.640×10⁻⁸m

Explanation:

Given data

Diameter d=78 in=1.9812 m

Cross Area is:

$A=(\pi /4)d^{2} \\A=(\pi /4)(1.9812m)^{2}\\A=3.08m^{2}$

Applied Load P=17 KN=17×10³N

E=29 × 106 psi=1.99947961×10¹¹Pa

Stress and Strain in x direction

Stress in x direction

σ=P/A

$=\frac{17*10^{3}N }{3.08m^{2} }\\ =5517.25Pa$

σ=5517.25 Pa

Strain in x direction

ε=σ/E

$=\frac{5517.25}{1.99947961*10^{11} } \\=2.76*10^{-8}$

ε=2.76×10⁻⁸

Part (a)

Elongation of the rod=Lε

=(0.2032)(2.76×10⁻⁸)

Elongation of the rod==5.61×10⁻⁹m

Part(b) Change in diameter

Strain in y direction

ε₁= -vε

ε₁= -(0.30)(2.76×10⁻⁸)

ε₁=-8.28×10⁻⁹

Change in diameter=d×ε₁

Change in diameter=(1.9812m)×(-8.28×10⁻⁹)

Change in diameter=1.640×10⁻⁸m

4 0

3 years ago

A particle with charge q = 10-9 C and mass m = 5.0 x 10-9 kg is moving in a magnetic field whose magnitude is 0.003 T. The speed

Marina86 [1]

Answer:

a=0.212 m/s²

Explanation:

Given that

q= 10⁻⁹ C

m = 5 x 10⁻⁹ kg

Magnetic filed ,B= 0.003 T

Speed ,V= 500 m/s

θ= 45°

Lets take acceleration of the mass is a m/s²

The force on the charge due to magnetic filed B

F= q V B sinθ

Also F= m a ( from Newton's law)

By balancing these above two forces

m a= q V B sinθ

$a=\dfrac{qVB\ sin\theta}{m}$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\ sin45^{\circ}}{5\times 10^{-9}}\ m/s^2$

$a=\dfrac{10^{-9}\times 500\times 0.003\times\dfrac{1}{\sqrt2}}{5\times 10^{-9}}\ m/s^2$

a=0.212 m/s²

4 0

3 years ago

A suspended object A is attracted to a neutral wall. It's also attracted to a positively-charged object B. Which of the followin

olga2289 [7]

There are two particular cases, the first is when Object A is attracted to the neutral wall. This would indicate that the object is not neutral, as there is an attraction.

At the same time we know that Object A is attracted to an object B. And therefore, the load of A must be opposite to that of B. Remember that opposite charges attract each other. If the charge of object B is positive, then the charge of object A will be negative.

Option B is correct: It has a negative charge.

7 0

3 years ago