I believe the answer would be protentional because they have the potential energy in them to lift the hammer.
The radius of the Sun is 6.96 x 108 m and the distance between the Sun and the Earth is roughtly 1.50 x 1011 m. You may assume that the Sun is a perfect sphere and that the irradiance arriving on the Earth is the value for AMO, 1,350 W/m2. Calculate the temperature at the surface of the Sun.
Answer:
5766.7 K
Explanation:
We are given that
Radius of Sun , R=[tex]6.96\times 10^{8} m[/tex]
Distance between the Sun and the Earth, D=[tex]1.50\times 10^{11}m[/tex]
Irradiance arriving on the Earth is the value for AMO=[tex]1350W/m^2[/tex]
We have to find the temperature at the surface of the Sun.
We know that
Temperature ,T=[tex](\frac{K_{sc}D^2}{\sigma R^2})^{\frac{1}{4}}[/tex]
Where [tex]K_{sc}=1350 W/m^2[/tex]
[tex]\sigma=5.67\times 10^{-8}watt/m^2k^4[/tex]
Using the formula
[tex]T=(\frac{1350\times (1.5\times 10^{11})^2}{5.67\times 10^{-8}\times (6.96\times 10^{8})^2})^{\frac{1}{4}}[/tex]
T=5766.7 K
Hence, the temperature at the surface of the sun=5766.7 K
A 2028 kg Oldsmobile traveling south on Abbott Road at 14.5 m/s is unable to stop on the ice covered intersection for a red light at Saginaw Street. The car collides with a 4146 kg truck hauling animal feed east on Saginaw at 9.7 m/s. The two vehicles remain locked together after the impact. Calculate the velocity of the wreckage immediately after the impact. Give the speed for your first answer and the compass heading for your second answer. (remember, the CAPA abbreviation for degrees is deg) -1.75
Answer:
v = 8.1 m/s
θ = -36.4º (36.4º South of East).
Explanation:
Assuming no external forces acting during the collision (due to the infinitesimal collision time) total momentum must be conserved.Since momentum is a vector, if we project it along two axes perpendicular each other, like the N-S axis (y-axis, positive aiming to the north) and W-E axis (x-axis, positive aiming to the east), momentum must be conserved for these components also.Since the collision is inelastic, we can write these two equations for the momentum conservation, for the x- and the y-axes:We can go with the x-axis first:[tex]p_{ox} = p_{fx} (1)[/tex]
⇒ [tex]m_{tr} * v_{tr}= (m_{olds} + m_{tr}) * v_{fx} (2)[/tex]
Replacing by the givens, we can find vfx as follows:[tex]v_{fx} = \frac{m_{tr}*v_{tr} }{(m_{tr} + m_{olds)} } = \frac{4146kg*9.7m/s}{2028kg+4146 kg} = 6.5 m/s (3)[/tex]
We can repeat the process for the y-axis:[tex]p_{oy} = p_{fy} (4)[/tex]
⇒[tex]m_{olds} * v_{olds}= (m_{olds} + m_{tr}) * v_{fy} (5)[/tex]
Replacing by the givens, we can find vfy as follows:[tex]v_{fy} = \frac{m_{olds}*v_{olds} }{(m_{tr} + m_{olds)} } = \frac{2028kg*(-14.5)m/s}{2028kg+4146 kg} = -4.8 m/s (6)[/tex]
The magnitude of the velocity vector of the wreckage immediately after the impact, can be found applying the Pythagorean Theorem to vfx and vfy, as follows:[tex]v_{f} = \sqrt{v_{fx} ^{2} +v_{fy} ^{2} }} = \sqrt{(6.5m/s)^{2} +(-4.8m/s)^{2}} = 8.1 m/s (7)[/tex]
In order to get the compass heading, we can apply the definition of tangent, as follows:[tex]\frac{v_{fy} }{v_{fx} } = tg \theta (8)[/tex]
⇒ tg θ = vfy/vfx = (-4.8m/s) / (6.5m/s) = -0.738 (9)
⇒ θ = tg⁻¹ (-0.738) = -36.4º
Since it's negative, it's counted clockwise from the positive x-axis, so this means that it's 36.4º South of East.A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m
Complete Question
A ball having mass 2 kg is connected by a string of length 2 m to a pivot point and held in place in a vertical position. A constant wind force of magnitude 13.2 N blows from left to right. Pivot Pivot F F (a) (b) H m m L L If the mass is released from the vertical position, what maximum height above its initial position will it attain? Assume that the string does not break in the process. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m.What will be the equilibrium height of the mass?
Answer:
[tex]H_m=1.65m[/tex]
[tex]H_E=1.16307m[/tex]
Explanation:
From the question we are told that
Mass of ball [tex]M=2kg[/tex]
Length of string [tex]L= 2m[/tex]
Wind force [tex]F=13.2N[/tex]
Generally the equation for [tex]\angle \theta[/tex] is mathematically given as
[tex]tan\theta=\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{F}{mg}[/tex]
[tex]\theta=tan^-^1\frac{13.2}{2*2}[/tex]
[tex]\theta=73.14\textdegree[/tex]
Max angle =[tex]2*\theta= 2*73.14=>146.28\textdegree[/tex]
Generally the equation for max Height [tex]H_m[/tex] is mathematically given as
[tex]H_m=L(1-cos146.28)[/tex]
[tex]H_m=0.9(1+0.8318)[/tex]
[tex]H_m=1.65m[/tex]
Generally the equation for Equilibrium Height [tex]H_E[/tex] is mathematically given as
[tex]H_E=L(1-cos73.14)[/tex]
[tex]H_E=0.9(1+0.2923)[/tex]
[tex]H_E=1.16307m[/tex]
According to Newton's law of universal gravitation, which statements are true?
As we move to higher altitudes, the force of gravity on us decreases.
O As we move to higher altitudes, the force of gravity on us increases,
O As we gain mass, the force of gravity on us decreases.
O Aswe gain mass, the force of gravity on us increases.
DAs we move faster, the force of gravity on us increases.
Energy from the Sun is transferred from the Earth’s surface to the atmosphere, resulting in
atmospheric convection currents that produce winds. How do physical properties of the air
contribute to convection currents?
a -The warmer air sinks because it is more dense than cooler air.
b -The warmer air rises because it is more dense than cooler air.
c- The warmer air sinks because it is less dense than cooler air.
d -The warmer air rises because it is less dense than cooler air.
He throws a second ball (B2) upward with the same initial velocity at the instant that the first ball is at the ceiling. c. How long after the second ball is thrown do the two balls pass each other? d. When the balls pass each other how far are they above the juggler’s hands? e. When they pass each other what are their velocities?
Answer:
hello your question has some missing parts
A juggler performs in a room whose ceiling is 3 m above the level of his hands. He throws a ball vertically upward so that it just reaches the ceiling.
answer : c) 0.39 sec
d) 2.25 m
e) 1.92 m/sec
Explanation:
The initial velocity of the first ball = 7.67 m/sec ( calculated )
Time required for first ball to reach ceiling = 0.78 secs ( calculated )
Determine how long after the second ball is thrown do the two balls pass each other
Distance travelled by first ball downwards when it meets second ball can be expressed as : d = 1/2 gt^2 = 9.8t^2 / 2
hence d = 4.9t^2 ----- ( 1 )
Initial speed of second ball = first ball initial speed = 7.67 m/sec
3 - d = 7.67t - 4.9t ---- ( 2 )
equating equation 1 and 2
3 = 7.67t therefore t = 0.39 sec
Determine how far the balls are above the Juggler's hands ( when the balls pass each other )
form equation 1 ;
d = 4.9 t^2 = 4.9 *(0.39)^2 = 0.75 m
therefore the height the balls are above the Juggler's hands is
3 - d = 3 - 0.75 = 2.25 m
determine their velocities when the pass each other
velocity = displacement / time
velocity = d / t = 0.75 / 0.39 sec = 1.92 m/sec
what is momentum of a train that is 60,000 kg that is moving at velocity of 17m/s?
explain your answer
Which of the following is a mixture?
a air
biron
Chydrogen
d nickel
Answer:
it will option option A hope it helps
A solid sphere of radius R = 5 cm is made of non-conducting material and carries a total negative charge Q = -12 C. The charge is uniformly distributed throughout the interior of the sphere.
What is the magnitude of the electric potential V at a distance r = 30 cm from the center of the sphere, given that the potential is zero at r = [infinity] ?
Answer:
V= -3.6*10⁻¹¹ V
Explanation:
Since the charge is uniformly distributed, outside the sphere, the electric field is radial (due to symmetry), so applying Gauss' Law to a spherical surface at r= 30 cm, we can write the following expression:[tex]E* A = \frac{Q}{\epsilon_{0} } (1)[/tex]
At r= 0.3 m the spherical surface can be written as follows:[tex]A = 4*\pi *r^{2} = 4*\pi *(0.3m)^{2} (2)[/tex]
Replacing (2) in (1) and solving for E, we have:[tex]E = \frac{Q}{4*\pi *\epsilon_{0}*r^{2} } = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)^{2} y} (3)[/tex]
Since V is the work done on the charge by the field, per unit charge, in this case, V is simply:V = E. r (4)Replacing (3) in (4), we get:[tex]V =E*r = E*(0.3m) = \frac{(9e9N*m2/C2)*(-12C)}{(0.3m)} = -3.6e11 V (5)[/tex]
V = -3.6*10¹¹ Volts.The electrical potential module will be [tex]-3.6*10^-^1^1 V[/tex]
We can arrive at this answer as follows:
To answer this, we owe Gauss's law. This is because the charge is evenly distributed across the sphere. This will be done as follows:[tex]E*A=\frac{Q}{^E0} \\\\\\A=4*\pi*r^2[/tex]
Solving these equations will have:[tex]E=\frac{Q}{4*\pi*^E0*r^2} \\E= \frac{(9e9N*m2/c2)*(-12C)}{(0.3m)^2y}[/tex]
As we can see, the electric potential is carried out on the field charge. In this case, using the previous equations, we can calculate the value of V as follows:[tex]V=E*r\\V=E*0.3m= \frac{(9e9N*m^2/C2)*(-12C)}{0.3m} \\V= -3.6*10^-^1^1 V.[/tex]
More information about Gauss' law at the link:
https://brainly.com/question/14705081
A 35 kg box initially sliding at 10 m/s on a rough surface is brought to rest by 25 N
of friction. What distance does the box slide?
Answer:
the distance moved by the box is 70.03 m.
Explanation:
Given;
mass of the box, m = 35 kg
initial velocity of the box, u = 10 m/s
frictional force, F = 25 N
Apply Newton's second law of motion to determine the deceleration of the box;
-F = ma
a = -F / m
a = (-25 ) / 35
a = -0.714 m/s²
The distance moved by the box is calculated as follows;
v² = u² + 2ad
where;
v is the final velocity of the box when it comes to rest = 0
0 = 10² + (2 x - 0.714)d
0 = 100 - 1.428d
1.428d = 100
d = 100 / 1.428
d = 70.03 m
Therefore, the distance moved by the box is 70.03 m.