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3 years ago

What are the units for volume and/or mass? Please answer and explain clearly thanks!

Physics

1 answer:

Paul [167]3 years ago

6 0

The units for mass are grams (g) and kilograms (kg)
the units for volume are millilitres (ml) and litres (l)

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Can we see a halo around the half moon?

Hatshy [7]

Answer:

No we cannot

Explanation:

But what causes a ring to appear around the moon? This phenomenon is called a "moon halo." According to the National Weather Service, this ring of light, which is actually an optical illusion, forms around the moon when moonlight refracts off ice crystals in cirrus clouds, high up in the Earth's atmosphere.

8 0

2 years ago

Read 2 more answers

A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave

damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance $x$ m by the time of collision, then train B would have traveled $(2881 - x)$ m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

$t_{A} = t_{B}$

$t_{A}$ is the time spent by train A

$t_{B}$ is the time spent by train B

From,

$Velocity = \frac{Distance }{Time }\\$

$Time = \frac{Distance}{Velocity}$

Since the time spent by the two trains is equal,

Then,

$\frac{Distance_{A} }{Velocity_{A} } = \frac{Distance_{B} }{Velocity_{B} }$

${Distance_{A} = x$ m

${Distance_{B} = 2881 - x$ m

${Velocity_{A} = 100$ m/s

${Velocity_{B} = 136$ m/s

Hence,

$\frac{x}{100} = \frac{2881 - x}{136}$

$136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\$

$x$ ≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0

3 years ago

Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons

dybincka [34]

<span>The electric force is given by:
F = [ k*(q1)*(q2) ] / d^2
F = Electric force
k = Coulomb's constant
q1 = Charge of one proton
q2 = Charge of second proton
d = Distance between centers of mass
Values:
F = unknown
k = 8.98E 9 N-m^2/C^2
q1 = 1.6E-19
q2 = 1.6E-19
d = 1.0E-15 m
Insert values into F = [ k*(q1)*(q2) ] / d^2
F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2
F = </span>229.888 N
answer
the electric force of repulsion between nuclear protons is 229.888 N

3 0

3 years ago

Read 2 more answers

An airplane accelerates from a speed of 88m/s to a speed of 132 m/s during a 15 second time interval. How far did the airplane t

Gelneren [198K]

Answer:

$1650\:\mathrm{m}$

Explanation:

We can use the following kinematics equations to solve this problem:

$v_f=v_i+at,\\{v_f}^2={v_i}^2+2a\Delta x$ .

Using the first one to solve for acceleration:

$132=88+a(15),\\15a=44,\\a=\frac{44}{15}=2.9\bar{3}\:\mathrm{m/s^2}$ .

Now we can use the second equation to solve for the distance travelled by the airplane:

$132^2=88^2+2\cdot2.9\bar{3}\cdot \Delta x,\\\Delta x= \frac{9680}{2\cdot2.9\bar{3}},\\\Delta x =\fbox{$ 1650\:\mathrm{m}$}$ (three significant figures).

6 0

3 years ago

Hi i have homework can you help me

soldier1979 [14.2K]

Explanation:

The particle will be at rest when its velocity $v(t)$ is equal to zero. Recall that the velocity is simply the derivative of the position $x(t)$ with respect to time:

$v(t) = \dfrac{dx}{dt}$

Since $x(t) = t^2 - 2t + 2$

then

$v(t) = \dfrac{d}{dt}(t^2 - 2t + 2) = 2t - 2 = 0$

Solving for t, we find that the particle will be at rest at

$t = 1\:\text{s}$

6 0

3 years ago