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3 years ago

The image formed by a convex spherical mirror will always be

2 answers:

7 0

a. inverted and larger than the object.

A mirror is used to see your own reflection. Earlier in the days, people would use a still water to see their reflection. It is basically used for grooming. Spherical mirrors have been used by earlier mathematicians and physicist to conduct experiments related to geometry.

ryzh [129]3 years ago

7 0

Penn Foster STudents: d. upright and smaller than the object.

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I am really struggling with this question because I can't find anything on aphelion and perihelion, it's not a topic we went ove

Hoochie [10]

I have a strange hunch that there's some more material or previous work
that goes along with this question, which you haven't included here.

I can't easily find the dates of Mercury's extremes, but here's some of the
other data you're looking for:

Distance at Aphelion (point in it's orbit that's farthest from the sun):
69,816,900 km
0. 466 697 AU

 Distance at Perihelion (point in it's orbit that's closest to the sun):
46,001,200 km
0.307 499 AU 

Perihelion and aphelion are always directly opposite each other in
the orbit, so the time between them is 1/2 of the orbital period.

Mercury's Orbital period = 87.9691 Earth days

1/2 (50%) of that is 43.9845 Earth days

The average of the aphelion and perihelion distances is

1/2 ( 69,816,900 + 46,001,200 ) = 57,909,050 km
or
1/2 ( 0.466697 + 0.307499) = 0.387 098 AU

This also happens to be 1/2 of the major axis of the elliptical orbit.

3 0

4 years ago

A roundabout is a type of playground equipment involving a large flat metal disk that is able to spin about its center axis. A r

Zina [86]

Answer:

116.1 kgm²/s

1.12718 rad/s

Decreases

Explanation:

m = Mass of girl = 43 kg

M = Mass of roundabout = 120 kg

v = Velocity of roundabout = 2.7 m/s

r = Radius of roundabout = 1 m = R

I = Moment of inertia

Her angular momentum

$L_i=mvr\\\Rightarrow L_i=43\times 2.7\times 1\\\Rightarrow L_i=116.1\ kgm^2/s$

Magnitude of angular momentum is 116.1 kgm²/s

Here the angular momentum is conserved

$L_f=L_i\\\Rightarrow I\omega=L_i\\\Rightarrow (\frac{1}{2}MR^2+mr^2)\omega=116.1\\\Rightarrow \omega=\frac{116.1}{\frac{1}{2}\times 120\times 1^2+43\times 1^2}\\\Rightarrow \omega=1.12718\ rad/s$

Angular speed of the roundabout is 1.12718 rad/s

Initial kinetic energy

$K_i=\frac{1}{2}mv^2\\\Rightarrow K_i=\frac{1}{2}43\times 2.7^2\\\Rightarrow K_i=156.735\ J$

Final kinetic energy

$K_f=\frac{1}{2}I\omega^2\\\Rightarrow K_f=\frac{1}{2}\times (\frac{1}{2}\times 120\times 1^2+43\times 1^2)\times 1.12718^2\\\Rightarrow K_f=65.43253\ J$

The overall kinetic energy decreases as can be seen. This loss is converted to heat.

5 0

3 years ago

What type of relationship does this graph show?

Basile [38]

Answer:

An nonlinear inverse relationship

Explanation:

7 0

3 years ago

A cart starts at x = +6.0 m and travels towards the origin with a constant speed of 2.0 m/s. What is it the exact cart position

Ira Lisetskai [31]

Answer:

At the origin (x' = 0 m)

Explanation:

Note: From the question, when the cart travels towards the origin, the magnitude of its exact position reduces with time.

The formula of speed is given as

S = d/t................. Equation 1

Where S = speed of the cart, d = distance covered by the cart over a certain time. t = time taken to cover the distance.

make d the subject of the equation,

d = St ................. Equation 2

Given: S = 2.0 m/s, t = 3.0 s

Substitute into equation 2

d = 2(3)

d = 6 m.

From the above, the cart covered a distance of 6 m in 3 s.

The exact position of the cart = Initial position-distance covered

x' = x-d............ Equation 3

Where x' = exact position of the cart 3 s later, x = initial position of the cart, d = distance covered by the cart in 3.0 s.

Given: x = +6.0 m, d = 6 m.

Substitute into equation 3

x' = +6-6

x' = 0 m.

Hence the cart will be at 0 m (origin) 3 s later

5 0

3 years ago

A man pushed a cabinet with a force of 200N. What is the mass of the cabinet that accelerates 4 m/s/s?

ELEN [110]

Answer:

$\boxed {\boxed {\sf 50 \ kg}}$

Explanation:

We are asked to find the mass of a cabinet, given the force and acceleration. According to Newton's Second Law of Motion, force is the product of mass and acceleration. The formula for this is:

$F= m \times a$

The force is 200 Newtons, but we should convert the units to make unit cancellation easier. 1 Newton is equal to 1 kilogram meter per second squared, so the force of 200 Newtons is 200 kilogram meters per second squared.

The mass is unknown and the acceleration is 4 meters per second per second or 4 meters per second squared.

F= 200 kg*m/s²
a= 4 m/s²

Substitute the values into the formula.

$200 \ kg *m/s^2 = m \times 4 \ m/s^2$

We are solving for the mass, m, so we must isolate the variable. It is being multiplied by 4 meters per second squared. The inverse operation of multiplication is division. Divide both sides by 4 m/s²

$\frac {200 \ kg *m/s^2}{4 \ m/s^2}= \frac{m \times 4\ m/s^2}{4 \ m/s^2}$

$\frac {200 \ kg *m/s^2}{4 \ m/s^2} =m$

The units of meters per second squared cancel.

$\frac {200 \ kg }{4 }=m$

$50 \ kg =m$

The mass of the cabinet is 50 kilograms.

5 0

3 years ago