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3 years ago

Sunspots. Which of the following is a true statement.

Physics

2 answers:

Mama L [17]3 years ago

7 0

Answer:

A). They are areas of the sun that are ''burnt out'' or the first option would be your answer.

Explanation:

i hope it helps someone

AlladinOne [14]3 years ago

6 0

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a golfer hits a golf ball with a velocity of 36.0 meters/second at an angle of 28.0. if the hang time of the golf ball is 33.4 s

Kobotan [32]

Solving this using the time, we know that range = horizontal velocity x time of flight

since there are no horizontal forces acting on the ball, there are no horizontal accelerations and the initial horizontal velocity of 36 cos 28 will be constant throughout. If we use the correct time of flight given the launch parameters, we have

range = 36 cos 28 x 3.44 s = 109.3 m

6 0

3 years ago

Two long wires hang vertically. Wire 1 carries an upward current of 1.50 A . Wire 2,20.0cm to the right of wire 1, carries a dow

Inessa [10]

The magnitude of the current in wire 3 is 2.4 A and in a direction pointing in the downward direction.

The force per unit length between two parallel thin current-carrying $I_1$ and $I_2$ wires at distance ' r ' is given by $f=\frac{u_0I_1I_2}{2\pi r}$ ....(1) .

If the current is flowing in both wires in the same direction, and the force between them will be the attractive force and if the current is flowing in opposite direction in wires then the force between them will be the repulsive force.

A schematic of the information provided in the question can be seen in the image attached below.

From the image, force on wire 2 due to wire 1 = force on wire 2 due to wire 3

$F_2_1=F_2_3$

Using equation (1) , we get

$\frac{u_0I_2I_1}{0.2} =\frac{u_0I_2I_3}{0.32} \\\\\frac{I_1}{0.2} =\frac{I_3}{0.32} \\\\\frac{1.50}{0.2} =\frac{I_3}{0.32} \\\\0.48=0.2I_3\\\\I_3=2.4A$

I₃ = 2.4 A and the current is pointing in the downward direction

Learn more about the magnitude and direction of forces here:

brainly.com/question/14879801?referrer=searchResults

#SPJ4

5 0

2 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

4 years ago

Body composition refers to:

NISA [10]

Bone, fat and muscle

3 0

3 years ago

Complete each statement about the sign of the work done on a baseball. Carlton catches a baseball and his hand moves backward as

dezoksy [38]

Answer and Explanation:

The work done is negative as the force is applied in the forward direction and the displacement of the hand is in the backward direction.
The work done is positive as the work is done on the baseball by Carlton. Moreover, both the force applied and the displacement of Carlton are in a straight line in the forward direction.
The work done in this case is zero as Carlton rides with constant speed, the Kinetic energy too remains constant and hence net work done is zero.
The work done in this case is positive as the work is done on the baseball by Carlton and as the velocity changes its Kinetic energy also varies wit it.
Also the force applied and the displacement of Carlton in this case are in the directed to the same path.

8 0

3 years ago