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2 years ago

15

What is 2f In concave mirrors?

1 answer:

musickatia [10]2 years ago

4 0

The object is farther away then two principles focal lengths from the concave lens

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Although he did not present a mechanism, what were the key points of Alfred Wegener’s proposal for the concept of continental dr

valentinak56 [21]

Answer: Alfred Wegener provided some of the important points that supported the theory of continental drift. They are as follows-

The continents were once all attached together, and this can be proved by studying the coastlines of some of the continents that perfectly matches with one another.
The appearance of similar rock types and similar fossils (including both animals and plants) has also contributed much information that continents were once all together.

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3 years ago

A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra

Tatiana [17]

Answer:

The centripetal acceleration of the car is $8\ m/s^2$ .

Explanation:

Let the mass of the car, $m=10^3\ kg$

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(20\ m/s)^2}{50\ m}$

$a=8\ m/s^2$

So, the centripetal acceleration of the car is $8\ m/s^2$ . Hence, this is the required solution.

4 0

3 years ago

Could anyone help me for this question please!! Really emergency!!!

kodGreya [7K]

16/9 m/s^2
negative 4/3 m/s^2
14 m/s
the last one is too detailed to do in my head while on the bus; sorry

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3 years ago

How do you calculate the gravitational force on a rubber ball which has a mass of 50g

AfilCa [17]

P (gravitational force) = m (mass) x g
<=> P = 0.05 x 10
<=> P = 0.5N

6 0

3 years ago

A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subj

KATRIN_1 [288]

Answer:

$\Delta t = 8 s$

Explanation:

As we know that the angular acceleration of the wheel due to friction is constant

so we can use kinematics

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

so we have

$(65 \times 2\pi) = (2\pi \times 9)(10) + \frac{1}{2}(\alpha)(10^2)$

$130\pi = 180\pi + 50 \alpha$

$\alpha = -\pi rad/s^2$

now time required to completely stop the wheel is given as

$\omega_f = \omega_i + \alpha t$

$0 = (2\pi \times 9) + (-\pi) t$

$t = 18 s$

now time required to stop the wheel is given as

$\Delta t = 18 - 10$

$\Delta t = 8 s$

6 0

2 years ago