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s344n2d4d5 [400]
4 years ago
8

For what position of the object will a spherical concave mirror project on the screen an image smaller than the object? a. betwe

en focus and center
b. between the focus and the mirror
c. at the center of curvature
d. beyond the center of curvature

Physics
1 answer:
luda_lava [24]4 years ago
3 0

Answer:

option (d)

Explanation:

A concave mirror always forms a real and inverted image of an object except when the object placed between pole and focus of the mirror.

When the object is placed beyond the centre of curvature, it forms a image which is smaller than the object but it is real and inverted in nature.

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In the Olympic shotput event, an athlete throws the shot with an initial speed of 13.0 m/s at a 47.0° angle from the horizontal.
MariettaO [177]

Answer:

18.7 m

Explanation:

v_{o} = initial speed of the shot = 13 m/s

θ = angle of launch from the horizontal = 47 deg

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 13 Sin47 = 9.5 m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 1.80 m

t = time of travel

Using the kinematics equation

y=v_{oy} t+(0.5)a_{y} t^{2}

- 1.80 = (9.5) t + (0.5) (- 9.8) t²

t = 2.11 s

Consider the motion along the horizontal direction

x = horizontal displacement of the shot

v_{ox} = initial velocity along horizontal direction = 13 Cos47 = 8.87 m/s

a_{x} = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.11 s

Using the kinematics equation

x=v_{ox} t+(0.5)a_{x} t^{2}

x = (8.87) (2.11) + (0.5) (0) (2.11)²

x = 18.7 m

8 0
3 years ago
5. A hollow cylinder of mass m, radius Rc, and moment of inertia I = mRc2 is pushed against a spring (with spring constant k) co
Makovka662 [10]

Explanation:

(a) Draw a free body diagram of the cylinder at the top of the loop.  At the minimum speed, the normal force is 0, so the only force is weight pulling down.

Sum of forces in the centripetal direction:

∑F = ma

mg = mv²/RL

v = √(g RL)

(b) Energy is conserved.

EE = KE + RE + PE

½ kd² = ½ mv² + ½ Iω² + mgh

kd² = mv² + Iω² + 2mgh

kd² = mv² + (m RC²) ω² + 2mg (2 RL)

kd² = mv² + m RC²ω² + 4mg RL

kd² = mv² + mv² + 4mg RL

kd² = 2mv² + 4mg RL

kd² = 2m (v² + 2g RL)

d² = 2m (v² + 2g RL) / k

d = √[2m (v² + 2g RL) / k]

8 0
3 years ago
A 500.0 kg elevator is pulled upward with a constant force of 5500.0 N for a distance of 50.0 m. What is the work done by the 55
igomit [66]

C. The work done by the 5500.0 N force on the elevator is 275,000 J.

<h3>Work done by the applied force</h3>

The work done by the 5500.0 N force on the elevator is calculated as follows;

W = Fd

where;

  • F is the applied force
  • d is the displacement

W = 5500 x 50

W = 275,000 J

Thus, the work done by the 5500.0 N force on the elevator is 275,000 J.

Learn more about work done here: brainly.com/question/8119756

#SPJ1

3 0
2 years ago
An electron and a proton are separated by a distance of 1.0 m. What happens to the force between them if the electron moves 0.5
r-ruslan [8.4K]

Explanation:

It is given that, an electron and a proton are separated by a distance of 1.0 m i.e d = 1 m . At this position, F is the force between them

F=k\dfrac{q_1q_2}{1^2}...............(1)

We need to find effect on force between them if the electron moves 0.5 m away from the proton. Let the force is F'.

F'=k\dfrac{q_1q_2}{0.5^2}...............(2)

On dividing equation (1) and (2) we get :

\dfrac{F}{F'}=\dfrac{k\dfrac{q_1q_2}{1^2}}{k\dfrac{q_1q_2}{0.5^2}}

\dfrac{F}{F'}=0.25

F' = 4 F

So, the distance between the electron and the proton is 0.5 m, the new force will be 4 times of previous force.

6 0
3 years ago
An object's acceleration is given by a(t)=a(t)=60t m/s260t m/s2 . if it begins at rest, how far has it gone after 10 seconds?
Alex Ar [27]
The object's acceleration is given by the expression: a(t) = 60t m/s^2. The distance it traveled after 10 seconds after starting from rest can be obtained by integrating the expression, a(t), twice in order to get the expression for the distance as a function of time. 

The first integral of a(t) is the velocity, v(t), of the object which is then equal to: v(t) = 30t^2. The integral of v(t) is then distance, d(t), of the object which is then equal to: d(t) = 10t^3. With the distance equation available, we simply plug in t = 10 seconds into the equation. It is then determined that the object has traveled 10,000 m or 10 km after 10 seconds starting from rest.    
6 0
3 years ago
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