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3 years ago

The Astronomer Carl Sagan is famous. Why?

1 answer:

6 0

Mainly because he was Johnny Carson's advisor and consultant
on space, astronomy, and science in general, and he appeared
on The Tonight Show Starring Johnny Carson many times.

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A circular loop of wire of cross-sectional area 0.12 m2 consists of 200 turns, each carrying 0.50 A. It is placed in a magnetic

defon

Answer:

0.52 Nm

Explanation:

A = 0.12 m^2, N = 200, i = 0.5 A, B = 0.050 T

Angle between the plane of loop and magnetic field = 30 Degree

Angle between the normal of loop and the magnetic field = 90 - 30 = 60 degree

θ = 60°

Torque = N i A B Sinθ

Torque = 200 x 0.5 x 0.12 x 0.050 x Sin 60

Torque = 0.52 Nm

4 0

3 years ago

A spring does 80 J of work launching a 1.85 kg rock into the air. Ignoring air resistance, how high will the rock go?

Svetlanka [38]

h=80/(1.85*9.8)=4.4 m

3 0

3 years ago

What is the main difference between velocity and speed?

Marina86 [1]

Speed is scalar, meaning it's only going to be like 74 mph, doesn't matter which direction. Velocity is a vector, meaning it has direction. You can go -74mph when talking about velocity, not speed.

4 0

3 years ago

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

6 0

3 years ago

A force of 55N accelerates a 7.5kg wagon at 5.3 m/s^2 along a road. How large is the frictional force?

Varvara68 [4.7K]

Answer:

Explanation:

A force of $55\text{ }N$ is acting on a wagon along the road. The wagon weights $7.5\text{ }kg$ . Acceleration of the wagon is given as $5.3\text{ }\frac{m}{s^{2}}$ .

Consider the block as the system, the forces acting are Frictional force, Gravitational force, Normal reaction and External force applied by us.

Gravitational Force and Normal Reaction cancel out each other.

Net External Force = Mass of system/wagon $\times$ Acceleration of wagon

$F_{ext}-F_{friction}=(7.5\text{ }kg)\times(5.3\text{ }\frac{m}{s^{2}})=39.75\text{ }N\\55\text{ }N-F_{friction}=39.75\text{ }N\\F_{friction}=15.25\text{ }N$

$F_{friction}$ has a negative sign because it opposes the motion of the wagon.

∴ Frictional Force = 15.25 N

4 0

3 years ago