Which describes one feature of the image formed by a convex mirror?????

Answers

Answer 1

Answer:

Answer:

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

Explanation:

The image formed by a convex mirror will always have its smaller than the size of the object no matter what the position of the object.

Also notice that convex mirror always makes virtual images.

Another feature of the convex mirror is that an upright image is always formed by the convex mirror.

An important mirror formula to remember which is applicable for both convex and mirrors

1/f= 1/u + 1/v

Here:

'u' is an object which gets placed in front of a spherical mirror of focal

length 'f' and image 'u' is formed by the mirror.

Answer 2

Answer:

Answer:

right side up

Explanation:

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Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?

Answers

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.
However, linear speeds of points at different distances from the center, are different.
Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:

$v = \omega*r (1)$

Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:

$\omega = (v_(out) )/(r_(out) ) = (11.5m/s)/(3.14m) = 3.7 rad/sec (2)$

As we have already said, ωout = ωin = 3.7 rad/sec

Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.
Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:
$\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)$

⇒ Δθ₁ = Δθ₂ = 18.5 rad.

The linear distance traveled by each child, will be related with the linear speed of them.
Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:

$v_(inn) = \omega * r_(inn) = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)$

vout is a given of the problem ⇒ vout = 11. 5 m/s

Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:

$d_(inn) = v_(inn) * t = 2.9m/s* 5.0 s = 14.5 m (5)$

$d_(out) = v_(out) * t = 11.5 m/s* 5.0 s = 57.5 m (6)$

The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:

$F_(c) = m*(v^(2))/(r) (7)$

Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:

$F_(cin) = m*(v_(in)^(2))/(r_(in)) = 25.4 kg* ((2.9m/s)^(2) )/(0.78m) = 273.9 N (8)$

In the same way, we get Fcout (the force on the boy near the outer edge):

$F_(cout) = m*(v_(out)^(2))/(r_(out)) = 25.4 kg* ((11.5m/s)^(2) )/(3.14m) = 1069.8 N (9)$

The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.
The maximum friction force is given by the product of the coefficient of static friction times the normal force.
Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.
As both boys have the same mass, the normal force is also equal.
This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:
$F_(frs) = \mu_(s) * m* g (10)$

If this force is greater than the centripetal force, the boy will be able to hold on.
So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.

A bullet with a mass of 20 g and a speed of 960 m/s strikes a block of wood of mass 4.5 kg resting on a horizontal surface. The bullet gets embedded in the block. The speed of the block immediately after the collision is:________. A) cannot be found because we don't know whether the surface is frictionless.
B) is 0.21 km/s.
C) is 65 m/s.
D) is 9.3 m/s.
E) None of these is correct

Answers

Answer:

4.25m/s

E. None of the option is correct

Explanation:

Using the law of conservation of momentum to solve the problem. According to the law, the sum of momentum of the bodies before collision is equal to the sum of the bodies after collision. The bodies move with the same velocity after collision.

Mathematically.

mu + MU = (m+M)v

m and M are the masses of the bullet and the block respectively

u and U are their respective velocities

v is their common velocity

from the question, the following parameters are given;

m = 20g = 0.02kg

u = 960m/s

M = 4.5kg

U =0m/s (block is at rest)

Substituting this values into the formula above to get v;

0.02(960)+4.5(0) = (0.02+4.5)v

19.2+0 = 4.52v

4.52v = 19.2

Dividing both sides by 4.52

4.52v/4.52 = 19.2/4.52

v = 4.25m/s

Since they have the same velocity after collision, then the speed of the block immediately after the collision is also 4.25m/s

Assume: The bullet penetrates into the block and stops due to its friction with the block.

The compound system of the block plus the

bullet rises to a height of 0.13 m along a

circular arc with a 0.23 m radius.

Assume: The entire track is frictionless.

A bullet with a m1 = 30 g mass is fired

horizontally into a block of wood with m2 =

4.2 kg mass.

The acceleration of gravity is 9.8 m/s2 .

Calculate the total energy of the composite

system at any time after the collision.

Answer in units of J.

Taking the same parameter values as those in

Part 1, determine the initial velocity of the

bullet.

Answer in units of m/s.

Answers

To solve this problem we will start considering the total energy of the system, which is given by gravitational potential energy of the total of the masses. So after the collision the system will have an energy equivalent to,

$E_T = (m_1+m_2)gh$

Here,

$m_1$ = mass of bullet

$m_2$ = Mass of Block of wood

The ascended height is 0.13m, so then we will have to

PART A)

$E_T = (m_1+m_2)gh$

$E_T = (0.03+4.2)(9.8)(0.13)$

$E_T = 5.389J$

PART B) At the same time the speed can be calculated through the concept provided by the conservation of momentum.

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Since the mass at the end of the impact becomes only one in the system, and the mass of the block has no initial velocity, the equation can be written as

$m_1v_1 =(m_1+m_2)v_f$

The final velocity can be calculated through the expression of kinetic energy, so

$E_T = KE = (1)/(2) (m_1+m_2)v_f^2$

$v_f = \sqrt{(2E_T)/(m_1+m_2)}$

$v_f = \sqrt{(2*5.389J)/(0.03+4.2)}$

$v_f = 1.5962m/s$

Using this value at the first equation we have that,

$m_1v_1 = (m_1+m_2)v_f$

$v_1 =((m_1+m_2)v_f)/(m_1)$

$v_1 = ((0.03+4.2)(1.5962))/(0.03)$

$v_1 = 225.06m/s$

As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None

Answers

Answer:

B- Velocity

Explanation:

This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

6. As distance increases, gravitational force *
(10 Points)
increases
decreases

Answers

It decreasessssssssss

A proton and an alpha particle (helium nucleus consisting of two protons and two neutrons) are accelerated from rest across the same potential difference. Assume the proton mass and the neutron mass are roughly the same and neglect any relativistic effect. Compared to the final speed of the proton, the final speed of the alpha particle is?1. less by a factor of 22. less by a factor of √ 23. less by a factor of 44. greater by a factor of 25. the same