Việt và Thái cùng nhảy ra khỏi một chiếc be với vận tốc bằng nhau. Việt có khối lượng 75kg và nhảy về hướng đông

What is the mass of an object that weigh 150 N on earth

aev [14]

450 g =0.9921 lb I hope this helps you out.

8 0

4 years ago

In the design of a supermarket, there are to be several ramps connecting different parts of the store. Customers will have to pu

tatuchka [14]

Answer:

3.90 degrees

Explanation:

Let g= 9.81 m/s2. The gravity of the 30kg grocery cart is

W = mg = 30*9.81 = 294.3 N

This gravity is split into 2 components on the ramp, 1 parallel and the other perpendicular to the ramp.

We can calculate the parallel one since it's the one that affects the force required to push up

F = WsinΘ

Since customer would not complain if the force is no more than 20N

F = 20

$294.3sin\theta = 20$

$sin\theta = 20/294.3 = 0.068$

$\theta = sin^{-1}0.068 = 0.068 rad = 0.068*180/\pi \approx 3.90^0$

So the ramp cannot be larger than 3.9 degrees

6 0

4 years ago

A ____ is the time required for one half of the nuclei in a radio- ____ isotope to decay.

sukhopar [10]

Answer:

A half-life is the time required for one half of the nuclei in a radio- active isotope to decay.

Explanation:

A radio-active isotope is an isotope which undergoes radioactive decay.

Radioactive decay is a spontaneous process in which the nucleus of an atom changes its state (turning into a different nucleus, or de-exciting), emitting radiation, which can be of three different types: alpha, beta or gamma.

The half-life of a radio-active isotope is the time required for half of the nuclei of the initial sample to decay.

The law of radio-active decay can be expressed as follows:

$N(t) = N_0 (\frac{1}{2})^{t/t_{1/2}}$

where

N(t) is the number of undecayed nuclei left at time t

N0 is the initial number of nuclei

t is the time

$t_{1/2}$ is the half-life

We see that when $t=t_{1/2}$ (that means, when 1 half-life has passed), the number of undecayed nuclei left is

$N(t) = N_0 (\frac{1}{2})^{t_{1/2}/t_{1/2}}=N_0 (\frac{1}{2})^1=\frac{N_0}{2}$

So, half of the initial nuclei.

5 0

3 years ago

25 PTS

yarga [219]

The answer would be C

7 0

4 years ago

Suppose that an asteroid traveling straight toward the center of the earth were to collide with our planet at the equator and bu

vlada-n [284]

Answer:

$\frac{1}{10}$ M

Explanation:

To apply the concept of angular momentum conservation, there should be no external torque before and after

As the asteroid is travelling directly towards the center of the Earth, after impact ,it does not impose any torque on earth's rotation, So angular momentum of earth is conserved

⇒ $I_{1} \times W_{1} =I_{2} \times W_{2}$

$I_{1}$ is the moment of interia of earth before impact
$W_{1}$ is the angular velocity of earth about an axis passing through the center of earth before impact
$I_{2}$ is moment of interia of earth and asteroid system
$W_{2}$ is the angular velocity of earth and asteroid system about the same axis

let $W_{1}=W$

since $\text{Time period of rotation}∝$ $\frac{1}{\text{Angular velocity}}$

⇒ if time period is to increase by 25%, which is $\frac{5}{4}$ times, the angular velocity decreases 25% which is $\frac{4}{5}$ times

therefore $W_{1}$ $=$ $\frac{4}{5} \times W_{1}$

$I_{1}=\frac{2}{5} \times M\times R^{2}$ (moment of inertia of solid sphere)

where M is mass of earth

R is radius of earth

$I_{2}=\frac{2}{5} \times M\times R^{2}+M_{1}\times R^{2}$

(As given asteroid is very small compared to earth, we assume it be a particle compared to earth, therefore by parallel axis theorem we find its moment of inertia with respect to axis)

where $M_{1}$ is mass of asteroid