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2 years ago

A 1000kg truck is moving at a velocity of 4m/s what is it's momentum?

Physics

1 answer:

Alona [7]2 years ago

6 0

Answer:

4000 kg m/s

Explanation:

p=mv

p = momentum in kg m/s
m = mass in kg
v = velocity in m/s

substitute values into the formula above to get:

p=1000x4

p=4000 kg m/s

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A distant planet with a mass of (7.2000x10^26) has a moon with a mass of (5.0000x10^23). The distance between the planet and the

BARSIC [14]

Answer:

Explanation:

This is a simple gravitational force problem using the equation:

$F_g=\frac{Gm_1m_2}{r^2}$ where F is the gravitational force, G is the universal gravitational constant, the m's are the masses of the2 objects, and r is the distance between the centers of the masses. I am going to state G to 3 sig fig's so that is the number of sig fig's we will have in our answer. If we are solving for the gravitational force, we can fill in everything else where it goes. Keep in mind that I am NOT rounding until the very end, even when I show some simplification before the final answer.

Filling in:

$F_g=\frac{(6.67*19^{-11})(7.2000*10^{26})(5.0000*10^{23})}{(6.10*10^{11})^2}$ I'm going to do the math on the top and then on the bottom and divide at the end.

$F_g=\frac{2.4012*10^{40}}{3.721*10^{23}}$ and now when I divide I will express my answer to the correct number of sig dig's:

$Fg=$ 6.45 × 10¹⁶ N

8 0

3 years ago

A train slows down as it rounds a sharp horizontal turn, going from 94.0 km/h to 46.0 km/h in the 17.0 s that it takes to round

Svetllana [295]

Answer:

1.41 m/s^2

Explanation:

First of all, let's convert the two speeds from km/h to m/s:

$u = 94.0 km/h \cdot \frac{1000 m/km}{3600 s/h} = 26.1 m/s$

$v=46.0 km/h \cdot \frac{1000 m/km}{3600 s/h}=12.8 m/s$

Now we find the centripetal acceleration which is given by

$a_c=\frac{v^2}{r}$

where

v = 12.8 m/s is the speed

r = 140 m is the radius of the curve

Substituting values, we find

$a_c=\frac{(12.8 m/s)^2}{140 m}=1.17 m/s^2$

we also have a tangential acceleration, which is given by

$a_t = \frac{v-u}{t}$

where

t = 17.0 s

Substituting values,

$a_t=\frac{12.8 m/s-26.1 m/s}{17.0 s}=-0.78 m/s^2$

The two components of the acceleration are perpendicular to each other, so we can find the resultant acceleration by using Pythagorean theorem:

$a=\sqrt{a_c^2+a_t^2}=\sqrt{(1.17 m/s^2)+(-0.78 m/s^2)}=1.41 m/s^2$

6 0

3 years ago

Read 2 more answers

Select the correct answer. Brian is repairing an old alarm clock. He needs to replace a device that converts the electric energy

Stella [2.4K]

it shouldbe a buzzer

3 0

2 years ago

Read 2 more answers

Read the paragraph below and answer the question that follows:

elena-s [515]

B. Helium because it is constantly being made in the sun

Explanation:

From the given paragraph, we can conclude that helium is the most common element in a star such as the sun because it is constantly being made.

In the nuclear fusion process that results in the production of helium, hydrogen nuclei are the reactants and helium is the product.

since the reactants are constantly being used in the core of the sun, this suggests that the products must be more.
The reaction requires a high temperature and pressure to initiate.
It is a series of chain reaction that uses a fuel of hydrogen to produce helium.

learn more:

Transmutation brainly.com/question/3433940

#learnwithBrainly

8 0

2 years ago

A solid, homogeneous sphere with a mass of m0, a radius of r0 and a density of rho0 is placed in a container of water. Initially

ivanzaharov [21]

Answer:

a) s,f,r b) r c) f

Explanation:

To determine what happens with the sphere we use Newton's second law with the Archimedes principle that states that the thrust (B) on a body is equal to the weight of the liquid dislodged

For the sphere to be in equilibrium the sum of forces is zero

B - W = 0

B = W = mg

Now let's use the concept of density for the body and water

Solid sphere

ρ = m / V

V = 4/3 π r³

m = ρ₀ (4/3 π r³)

W = ρ₀ (4/3 π r³) g

Water (a)

ρ = mₐ / Vₐ

mₐ = ρ Vₐ

B = ρ Vₐ g

Let's replace and simplify

ρ Vₐ g = ρ₀ (4/3 π r³) g

ρ Vf = ρ₀ (4/3 π r³) (1)

For the initial condition with rho, mo and ro the height of the water is H, let's analyze each case

a) We have the same mass, but less radius, as density is mass over volume density increases

r <ro V <V₀ ⇒ ρ₁> ρ₀

When analyzing the equation (1) on the right side, this case is the most complicated because I can make the relationship between the density of the sphere and its volume change even when the mass is constant

Assume the three possibilities

- The product of (ρ₁ V) that does not matter in that case the left side does not change and the mark remains the same (s)

- The product (ρ₁ V) increases the left side must increase so the mark goes up (r)

- The product (ρ₁ V) decreases the left side should go down, so the low mark (f)

b) sphere the same radius, but the density increases.

In this case the right side of the equation (1) increases, therefore the left side must increase so that the volume must increase and consequently increase the height (r)

c) you have the same radius, but the mass decreases

r = r₀ V = V₀ m <m₀ ρ₁ <ρ₀

The right side of the equation decreases, because the density decreases, the left side must decrease, for this the volume must decrease, lowering the height (f)

8 0

2 years ago