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2 years ago

A flat, square surface with side length 4.35 cm is in the xy-plane at z=0.

Physics

1 answer:

Hunter-Best [27]2 years ago

4 0

Answer:

I X j = k (perpendicular to plane)

j X k = 0

i X k = 0 cross products for vectors i, j, k

Flux = B dot A and only B in the k direction is relevant

Flux = .550 T * .0435^2 m^2 = .00104 Webers-m^2

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How do you calculate change in position? A. initial position times two B. final position plus initial position C. final position

e-lub [12.9K]

The answer is C. Final position minus initial position.

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3 years ago

Which objects will likely have the smallest gravitational force between them?

DENIUS [597]

Answer:

A. Two tennis balls that are near each other

Explanation:

The formula for gravitational force (F) between two objects is

$F = \dfrac{Gm_{1}m_{2}}{d^{2} }$

where m₁ and m₂ are the masses of the two objects, d is the distance between their centres, and G is the gravitational constant.

Thus, two objects that are far from each other will have a smaller gravitational force. We can eliminate Options C and D.

If the objects are at the same distance, those with the smaller mass will have a smaller force.

The mass of a tennis ball is 57 g.

The mass of a soccer ball is 430 g.

Two tennis balls that are near each other will have a smaller gravitational attraction.

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3 years ago

A ship sets out to sail to a point 120 km due north. an unexpected storm blows the ship to a point 100 km due east of its starti

Pavel [41]

If you draw the problem, it would look like that shown in the attached picture. The total length the ship will now travel can be solved using the Pythagorean theorem. The solution is as follows:

d = √(120)²+(100)²
d = 156.2 km

So, the ship will have to travel 156.2 km to the northwest direction.

8 0

3 years ago

A 50.0-g object connected to a spring with a force constant of 35.0 n/m oscillates with an amplitude of 4.00 cm on a frictionles

Dimas [21]

A) The total energy of the system is sum of kinetic energy and elastic potential energy:
$E=K+U= \frac{1}{2}mv^2 + \frac{1}{2}kx^2$
where
m is the mass
v is the speed
k is the spring constant
x is the elongation/compression of the spring

The total energy is conserved, so we can calculate its value at any point of the motion. If we take the point of maximum displacement:
$x=A=4.00 cm = 0.04 m$
then the velocity of the system is zero, so the total energy is just potential energy, and it is equal to
$E=U= \frac{1}{2}kA^2 = \frac{1}{2}(35.0 N/m)(0.04 m)^2=0.028 J$

b) When the position of the object is
$x=1.00 cm = 0.01 m$
the potential energy of the system is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.01 m)^2 = 1.75 \cdot 10^{-3} J$
and so the kinetic energy is
$K=E-U=0.028 J - 1.75 \cdot 10^{-3}J =0.026 J$
since the mass is $m=50.0 g=0.05 kg$ , and the kinetic energy is given by
$K= \frac{1}{2}mv^2$
we can re-arrange the formula to find the speed of the object:
$v= \sqrt{ \frac{2K}{m} }= \sqrt{ \frac{2 \cdot 0.026 J}{0.05 kg} }=1.02 m/s$

c) The potential energy when the object is at
$x=3.00 cm=0.03 m$
is
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$
Therefore the kinetic energy is
$K=E-U=0.028 J-0.016 J = 0.012 J$

d) We already found the potential energy at point c, and it is given by
$U= \frac{1}{2}kx^2 = \frac{1}{2}(35.0 N/m)(0.03 m)^2 =0.016 J$

5 0

3 years ago

What is the length of the shadow cast on the vertical screen by your 10.0 cm hand if it is held at an angle of θ=30.0∘ above hor

Alchen [17]

Answer:

The length is $D = 5 \ cm$

Explanation:

From the question we are told that

The length of the hand is $l = 10.0 \ cm$

The angle at the hand is held is $\theta = 30 ^o$

Generally resolving the length the length of the hand to it vertical component we obtain that the length of the shadow on the vertical wall is mathematically evaluated as

$D = l * sin(\theta )$

substituting values

$D = 10 * sin (30)$

$D = 5 \ cm$

5 0

3 years ago