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Fittoniya [83]
3 years ago
10

What is concave mirror​

Physics
1 answer:
Licemer1 [7]3 years ago
8 0

Answer:

A curved mirror is a mirror with a curved reflecting surface. The surface may be either convex or concave. Most curved mirrors have surfaces that are shaped like part of a sphere, but other shapes are sometimes used in optical devices.

Explanation:

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In a very busy off-campus eatery one chef sends a 235-g broccoli-tomato-pickle-onion-mushroom pizza sliding down the counter fro
STALIN [3.7K]

Answer:

0.47922 kgm/s and will move right to left

Explanation:

m_1 = Mass of broccoli-tomato-pickle-onion-mushroom pizza = 235 g

m_2 = Mass of veggieburger-on-a-bun = 351 g

v_1 = Velocity of broccoli-tomato-pickle-onion-mushroom pizza = 1.65 m/s

v_2 = Velocty of veggieburger-on-a-bun = -2.47 m/s

Left to right is considered positive direction and the opposite is negative direction

Net momentum is

p=m_1v_1+m_2v_2\\\Rightarrow p=0.235\times 1.65+0.351\times -2.47\\\Rightarrow p=-0.47922\ kgm/s

The combined dish has a momentum of 0.47922 kgm/s and will move right to left

3 0
3 years ago
Two point charges of 30 nC and – 40 nC are held fixed on an x-axis at the origin and
love history [14]

the Ans is=

mp= 22.10^-6 kg

I hope it is helpful...

plzz mark it as brainliest

6 0
3 years ago
A basketball star covers 2.70 m horizontally in a jump to dunk the ball (see figure). His motion through space can be modeled pr
Likurg_2 [28]

Answer:

Part a)

T = 0.81 s

Part b)

v_x = 3.33 m/s

Part c)

v_y = 3.91 m/s

Part d)

\theta = 49.55 degree

Part e)

T = 1.11 s

Explanation:

Part a)

initial vertical position = 1.02 m

maximum height = 1.80 m

\Delta y = 1.80 - 1.02

\Delta y = 0.78 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(0.78)

v_y = 3.91 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 3.91 - 9.81 t_1

t_1 = 0.39 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(1.80 - 0.95)

v_y = 4.08 m/s

time taken by it to reach this height

4.08 = v_i + at

4.08 = 0 + 9.81 t_2

t_2 = 0.42 s

T = t_1 + t_2

T = 0.81 s

Part b)

Horizontal velocity

v_x = \frac{x}{t}

v_x = \frac{2.70}{0.81}

v_x = 3.33 m/s

Part c)

vertical velocity is the intial y direction velocity

v_y = 3.91 m/s

Part d)

Take off angle is given as

tan\theta = \frac{3.91}{3.33}

\theta = 49.55 degree

Part e)

initial vertical position = 1.20 m

maximum height = 2.50 m

\Delta y = 2.50 - 1.20

\Delta y = 1.30 m

v_f^2 - v_y^2 = 2a \Delta y

0 - v_y^2 = 2(-9.81)(1.30)

v_y = 5.05 m/s

time taken by it to reach this height

v_y = v_i + at

0 = 5.05 - 9.81 t_1

t_1 = 0.51 s

Now when it again touch the ground then its speed is given as

v_f^2 - v_y^2 = 2a \Delta y

v_f^2 - 0 = 2(9.81)(2.50 - 0.72)

v_y = 5.9 m/s

time taken by it to reach this height

5.9 = v_i + at

5.9 = 0 + 9.81 t_2

t_2 = 0.60 s

T = t_1 + t_2

T = 1.11 s

5 0
3 years ago
Help mee pleaseee :)))
Anettt [7]

Answer:

See the explanation below.

Explanation:

Solving the first image question:

C ) The resulting force is defined by Newton's second law which tells us that the sum of the forces on a body is equal to the product of mass by acceleration. That is, there must be a force that acts on a body to produce an acceleration. If there is no acceleration it is because there are no external forces or developed by the body. And if there is no acceleration the body moves at a constant speed, in a straight line, so the response is C.

For the second image, we must remember that weight is defined as the product of mass by gravitational acceleration.

W = m*g

where:

W = weight [N]

m = mass [kg]

g = gravity acceleration [m/s²]

Now we have

m = 50 [kg]

ge = Earth gravity acceleration = 10 [m/s²]

gp = Distant planet gravity acceleration = 4 [m/s²]

We = ge*m

We = 10*50 = 500 [N]

Wp =gp*m

Wp = 4*50 = 200 [N]

Therefore the answer is D

For the third image, The mass is always going to be preserved, regardless of where the body or object is in space, its weight is the only one that changes since the gravitational force is modified. That is, the mass on the moon and on Earth will always be the same.

m = 70 [kg]

First, we must calculate the acceleration, by means of the following equation of kinematics.

v_{f} =v_{o} +a*t

where:

Vf = final velocity = 20 [m/s]

Vo = initial velocity = 0 (because stars from the rest)

a = acceleration [m/s²]

t = time = 4 [s]

20 = 0 + a*4

20 = 4*a

a = 5 [m/s²]

Now using Newton's second law which tells us that the total force acting on a body is equal to the product of mass by acceleration.

F = m*a

where:

F = force [N] (units of Newtons)

m = mass = 2 [kg]

a = acceleration = 5 [m/s²]

F = 2*5

F = 10 [N]

The body of Figure D, since a total force of 25 [N] to the left acts on it, in the rest of cases the force is zero or much less than 25 [N]

50 + 40 - 35 - 30 = F

F = 25 [N]

8 0
3 years ago
Spiderman, whose mass is 74.0 kg, is dangling on the free end of a 11.0-m-long rope, the other end of which is fixed to a tree l
Anestetic [448]

Answer:

W = -1844.513 J

Explanation:

GIVEN DATA:

mass of spider man is m  74 kg

vertical displacement if spider is 11 m

final displacement  =  11 cos 60.6 =  - 6.753 m

change in displacement is  = -6.753 - (-11) = 4.25 m

gravity force act on spiderman is f = mg = 74 × 9.8 = 725.2 N

work done by gravity is W = F \delta r cos\theta

W = 725.2 \times 4.25 \times cos 180

where 180 is the angle between spiderman weight and displacement

W = -1844.513 J

7 0
3 years ago
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