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3 years ago

The Tri-Rail traveled a distance of 8,000 m at an

Physics

1 answer:

Anton [14]3 years ago

8 0

Answer:

Time taken = 444.4s

Explanation:

how long = the time taken

and from the definition of speed,we have that :

Average speed = distance ( m)

_____________

time (s)

therefore,Time = distance (m) / average speed (m/s)

Time = 8000/18

Time = 444.4 (s)

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A flatbed truck is carrying a 20.0 kg crate along a level road. the coefficient of static friction between the crate and the bed

Dahasolnce [82]

<span>3.92 m/s^2 Assuming that the local gravitational acceleration is 9.8 m/s^2, then the maximum acceleration that the truck can have is the coefficient of static friction multiplied by the local gravitational acceleration, so 0.4 * 9.8 m/s^2 = 3.92 m/s^2 If you want the more complicated answer, the normal force that the crate exerts is it's mass times the local gravitational acceleration, so 20.0 kg * 9.8 m/s^2 = 196 kg*m/s^2 = 196 N Multiply by the coefficient of static friction, giving 196 N * 0.4 = 78.4 N So we need to apply 78.4 N of force to start the crate moving. Let's divide by the crate's mass 78.4 N / 20.0 kg = 78.4 kg*m/s^2 / 20.0 kg = 3.92 m/s^2 And you get the same result.</span>

6 0

3 years ago

A crate of eggs is located in the middle of the flatbed of a pickup truck as the truck negotiates a curve in the flat road. The

oksian1 [2.3K]

Answer:

v = 14.35 m/s

Explanation:

As we know that crate is placed on rough bed

so here when pickup will take a turn around a circle then in that case the friction force on the crate will provide the necessary centripetal force on the crate

So here we have

$\mu mg = \frac{mv^2}{R}$

here we have

$\mu g = \frac{v^2}{R}$

now we know that

$v = \sqrt{\mu Rg}$

here we have

$\mu = 0.600$

R = 35 m

g = 9.81 m/s/s

now plug in all values in above equation

$v = \sqrt{(0.600)(35)(9.81)}$

$v = 14.35 m/s$

4 0

3 years ago

Which statement describes the relationship of resistance and current

slavikrds [6]

This is picture describe as

More current mean less resistance

Less resistance mean more current

6 0

3 years ago

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli

Pie

Answer:

a) v = 3,843 m / s, b) 46.7º North- East

Explanation:

Moment is a vector quantity, so one of the best ways to solve this problem is to solve each component separately.

The system is formed by the two vehicles so that the moment is preserved during the crash

Direction to the East

initial instant. Before the crash

p₀ = mₐ vₐ₀

final insttne. After the crash

p_f = (mₐ + m_b) vₓ

p₀ = p_f

mₐ vₐ₀ = (mₐ + m_b) vₓ

vₓ = $\frac{m_a}{m_a + m_b} \ v_{ao}$

let's calculate

vₓ = $\frac{16.7}{16.7 + 29.3} \ 7.26$

vₓ = 2,636 m / s

direction north

initial p₀ = m_b v_{bo}

final p_f = (mₐ + m_b) v_y

p₀ = p_f

m_b v_{bo} = (mₐ + m_b) v_y

v_y = $\frac{m_b}{m_a+m_b} \ v_{bo}$

let's calculate

v_y = $\frac{29.3}{16.7 + 29.3} \ 4.39$

v_y = 2.796 m / s

the final speed of the two two vehicles is

v = (2,636 i ^ + 2,796 j ^) m / s

a) the magnitude of the velocity

let's use the Pythagorean theorem

v = $\sqrt{v_x^2 + v_y^2}$

v = $\sqrt{2.636^2 + 2.796^2}$

v = 3,843 m / s

b) let's use trigonometry to find the direction

tan θ = v_y / vₓ

θ = tan⁻¹ v_y / vₓ

θ = tan⁻¹ (2,796 / 2,636)

θ = 46.7º

This direction is 46.7º North East

7 0

3 years ago

A sample of nitrogen occupies 5.50 liters under a pressure of 900 torr at 25oC. At what temperature will it occupy 10.0 liters a

vitfil [10]

Answer:

<u>At 268.82°C</u> volume occupied by nitrogen is 10 liters at pressure of 900 torr.

Explanation:

Given:

Volume of a sample of nitrogen = 5.50 liters

Pressure = 900 torr

Temperature = 25°C

To find the temperature at which the nitrogen will occupy 10 liters volume at same pressure.

Solution:

Since the pressure is kept constant, so we can apply the temperature-volume law also called the Charles Law.

Charles Law states that the volume of a gas held at constant pressure is directly proportional to the temperature of the gas in Kelvin.

Thus, we have :

$V$ ∝ $T$

$\frac{V}{T}=k$

where $k$ is a constant.

For two samples of gases, the law can be given as:

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

From the data given:

$V_1=5.5\ l$

$T_1=25\ \°C =(273+25)K= 298 K$

$V_2=10\ l$

We need to find $T_2$ .

Plugging in values in the formula.

$\frac{5.5}{298}=\frac{10}{T_2}$

Multiplying both sides by $T_2$ .

$T_2\times\frac{5.5}{298}=\frac{10}{T_2}\times T_2$

$\frac{5.5}{298}T_2={10}$

Multiplying both sides by $\frac{298}{5.5}$

$\frac{298}{5.5}\times\frac{5.5}{298}T_2=\frac{10\times 298}{5.5}$

$T_2=541.82\ K$

$T_2=541.82\ K-273\ K = 268.82\°C$

Thus, at 268.82°C volume occupied by nitrogen is 10 liters at pressure of 900 torr.

7 0

3 years ago