If an object is moving with a constant acceleration, then it could NOT be

A person gives a shopping cart an initial push along a horizontal floor to get it moving, and then lets go. The cart travels for

mash [69]

Answer:

Only a backward force is acting, no forward force.

Explanation:

Once released from the initial push, in absence of friction, the shopping cart would continue moving forward at a constant speed forever.
As it would move at a constant speed, no net force would be acting on it.
So, if it is gradually slowing, there must be a net force producing an acceleration in a direction opposite to the movement.
This force is the kinetic friction force, and is the only force acting on the cart in the horizontal direction.
As any friction force, opposes to the relative movement between the cart and the horizontal floor, which means that is directed backward.
This is consistent with the direction of the acceleration of the cart.

8 0

3 years ago

Suppose a small planet is discovered that is 16 times as far from the Sun as the Earth's distance is from the Sun. Use Kepler's

mamaluj [8]

Answer:

23376 days

Explanation:

The problem can be solved using Kepler's third law of planetary motion which states that the square of the period T of a planet round the sun is directly proportional to the cube of its mean distance R from the sun.

$T^2\alpha R^3\\T^2=kR^3.......................(1)$

where k is a constant.

From equation (1) we can deduce that the ratio of the square of the period of a planet to the cube of its mean distance from the sun is a constant.

$\frac{T^2}{R^3}=k.......................(2)$

Let the orbital period of the earth be $T_e$ and its mean distance of from the sun be $R_e$ .

Also let the orbital period of the planet be $T_p$ and its mean distance from the sun be $R_p$ .

Equation (2) therefore implies the following;

$\frac{T_e^2}{R_e^3}=\frac{T_p^2}{R_p^3}....................(3)$

We make the period of the planet $T_p$ the subject of formula as follows;

$T_p^2=\frac{T_e^2R_p^3}{R_e^3}\\T_p=\sqrt{\frac{T_e^2R_p^3}{R_e^3}\\}................(4)$

But recall that from the problem stated, the mean distance of the planet from the sun is 16 times that of the earth, so therefore

$R_p=16R_e...............(5)$

Substituting equation (5) into (4), we obtain the following;

$T_p=\sqrt{\frac{T_e^2(16R_e)^3}{(R_e^3}\\}\\T_p=\sqrt{\frac{T_e^24096R_e^3}{R_e^3}\\}$

$R_e^3$ cancels out and we are left with the following;

$T_p=\sqrt{4096T_e^2}\\T_p=64T_e..............(6)$

Recall that the orbital period of the earth is about 365.25 days, hence;

$T_p=64*365.25\\T_p=23376days$

4 0

3 years ago

How does the structure of the stigma aid in pollination

natali 33 [55]

<span>Pollination is the process by which pollen is transferred to the female part of a plant. The stigma is the central part of the flower, that is supported by a style and is part of the female reproductive organ within plants. Its structure is optimised to promote pollination by having hairs, sticky surfaces and three-dimensional sculptures that capture and trap pollen.</span>

5 0

3 years ago

. Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connecte

Umnica [9.8K]

Given :

Reem took a wire of length 10 cm. Her friend Nain took a wire of 5 cm of the same material and thickness both of them connected with wires as shown in the circuit given in figure. The current flowing in both the circuits is the same.

To Find :

Will the heat produced in both the cases be equal.

Solution :

Heat released is given by :

H = i²Rt

Here, R is resistance and is given by :

$R = \dfrac{\rho L}{A}$

So,

$H = i^2\times \dfrac{\rho L}{A} t\\\\H = \dfrac{i^2\rho Lt}{A}$

Now, in the question every thing is constant except for the length of the wire and from above equation heat is directly proportional to the length of the wire.

So, heat produced by Reem's wire is more than Nain one.

Hence, this is the required solution.

7 0

3 years ago

Find the magnitude of the electric field due to a charged ring of radius "a" and total charge "Q", at a point on the ring axis a

34kurt

Answer:

E= $\frac{KQ}{2\sqrt 2a^2}$