What do you do if you approach a railroad crossing with a crossbuck sign that has no lights or gates

the force between a planet and a spacecraft is 1 million newtons. whst will the force be iv the spacecraft moves to halfg its or

dybincka [34]

The forces of gravity between two objects are inversely proportional to
the square of the distance between them. So reducing the distance
by 1/2 means increasing the gravitational force by 2² = 4 times.
The 1 million newtons becomes 4 million newtons.

Note that this does NOT mean the satellite's altitude above the surface.
When you're calculating gravitational forces, it's the distance between
the centers of the objects. So the question is a meaningful exercise
only if we use the distance between the satellite and the planet's center.

6 0

3 years ago

What is terminal velocity and when is it reached?

Ne4ueva [31]

Terminal velocity is the velocity at which a falling body experience when its weight is equal to the force resistance of force opposing the fall.

At terminal velocity the acceleration of the body is zero, which implies that the value of the this velocity is a uniform.

8 0

3 years ago

I need to write an acrostic poem on Globalisation and i need help. Does anyone have any ideas?

Aleonysh [2.5K]

Answer:

you could talk about the people and say something like change

Explanation:

5 0

3 years ago

How many photons will be required to raise the temperature of 1.8 g of water by 2.5 k ?'?

tatyana61 [14]

Missing part in the text of the problem:
"<span>Water is exposed to infrared radiation of wavelength 3.0×10^−6 m"</span>

First we can calculate the amount of energy needed to raise the temperature of the water, which is given by
$Q=m C_s \Delta T$
where
m=1.8 g is the mass of the water
$C_s = 4.18 J/(g K)$ is the specific heat capacity of the water
$\Delta T=2.5 K$ is the increase in temperature.

Substituting the data, we find
$Q=(1.8 g)(4.18 J/(gK))(2.5 K)=18.8 J=E$

We know that each photon carries an energy of
$E_1 = hf$
where h is the Planck constant and f the frequency of the photon. Using the wavelength, we can find the photon frequency:
$\lambda = \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{3 \cdot 10^{-6} m}=1 \cdot 10^{14}Hz$

So, the energy of a single photon of this frequency is
$E_1 = hf =(6.6 \cdot 10^{-34} J)(1 \cdot 10^{14} Hz)=6.6 \cdot 10^{-20} J$

and the number of photons needed is the total energy needed divided by the energy of a single photon:
$N= \frac{E}{E_1}= \frac{18.8 J}{6.6 \cdot 10^{-20} J} =2.84 \cdot 10^{20} photons$

4 0

3 years ago

Along a horizontal snow-covered track, a sled, of mass m = 105 kg, slides by the action of a horizontal force of 230 N. The coef

Andrew [12]

Answer:

Explanation:

The only thing I can figure you need here is the accleration of the sled. The equation we need to find this is Newton's Second Law that says that sum of the forces acting on an object is equal to the object's mass times its acceleration. For us, that looks like this because of the friction working against the sled:

F - f = ma but of course it's much more involved than that simple equation! We have the F value as 230 N, and we have the mass as 105, but we do not have the frictional force, f, and we need it to solve for a in the above equation. We know that

f = μ $F_n$ where μ is the coefficient of friction, and $F_n$ is the normal force, aka weight of the object. We will use the coefficient of friction and find the weight in order to fill in for f:

$F_n=mg$ so

$F_n=(105)(9.8)$ so the weight of the sled is

$F_n=$ 1.0 × 10³ with the correct number of sig dig there. Now to find f:

f = (.025)(1.0 × 10³) so

f = 25 to the correct number of sig fig. Now on to our "real" equation:

F - f = ma and

230 - 25 = 105a. We have to do the subtraction first, round, and then divide since the rules for addition and subtraction are different from the rules for dividing and multiplying.

230 - 25 will round to the tens place giving us 210. Then

210 = 105a. 210 has 2 sig figs in it while 105 has 3, so we will divide and round to 2 sig fig:

a = 2.0 m/sec²

3 0

3 years ago