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3 years ago

If your average speed is 3 m/s, how far have you traveled in 1 second, 2 second, 3 seconds?

Physics

2 answers:

yulyashka [42]3 years ago

8 0

1 second: 3 miles, 2 seconds: 6 miles 3 seconds: 9 miles

vichka [17]3 years ago

4 0

Answer:

in 1 second 3m, in 2 seconds 6m, in 3 seconds 9m.

Explanation:

distance=speed × time

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water flows horizontally from a hose that is 3m above the ground. the water lands 7.2m to the right of the hose . how long does

Sidana [21]

Answer:

0.782 s

Explanation:

The water flows horizontally from the hose, so its initial vertical velocity is 0.

Given:

y₀ = 3 m

y = 0 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 3 m + (0 m/s) t + ½ (-9.8 m/s²) t²

t = 0.782 s

Round as needed.

8 0

3 years ago

Read 2 more answers

Starting from rest, a 2-m-long pendulum swings from an angleof

Andrews [41]

Answer:

D.) 1m/s

Explanation:

Assume the initial angle of the swing is 12.8 degree with respect to the vertical. We can calculate the vertical distance from this initial point to the lowest point by first calculate the vertical distance from this point the the pivot point:

$L_1 = L*cos(12.8) = 2*0.975 = 1.95 m$

where L is the pendulum length

The vertical distance from the lowest point to the pivot point $L_2$ is the pendulum length 2m

this means the vertical distance from this initial point to the lowest point is simply:

$L_3 = L_2 - L_1 = 2 - 1.95 = 0.05 m$

As the pendulum travel (vertically) from the initial point to the bottom point, its potential energy is converted to kinetic energy:

$E_p = E_k$

$mgh = mv^2/2$

where m is the mass of the pendulum, g = 10 m/s2 is the constant gravitational acceleration, h = 0.05 is the vertical it travels, v is the pendulum velocity at the bottom, which we are trying to solve for.

The m on both sides of the equation cancel out

$v^2 = 2gh = 2*10*0.05 = 1$

$v = \sqrt{1} = 1 m/s$

so D is the correct answer

5 0

3 years ago

State pascal law and write three instrument which base unit?

OLEGan [10]

Answer:

Pascal's law (also Pascal's principle[1][2][3] or the principle of transmission of fluid-pressure) is a principle in fluid mechanics given by Blaise Pascal that states that a pressure change at any point in a confined incompressible fluid is transmitted throughout the fluid such that the same change occurs everywhere.[4] The law was established by French mathematician Blaise Pascal in 1653 and published in 1663.[5][6]

7 0

3 years ago

Need help ASAP shoving brainlest plsssss

seraphim [82]

The answer for this would be B!!

3 0

3 years ago

*please refer to photo* An electric field of magnitude 5.25 ✕ 10^5N/C points due south at a certain location. Find the magnitude

kvv77 [185]

Answer:

Approximately $3.86\; {\rm N}$ (given that the magnitude of this charge is $-7.35\; {\rm \mu C}$ .)

Explanation:

If a charge of magnitude $q$ is placed in an electric field of magnitude $E$ , the magnitude of the electrostatic force on that charge would be $F = E\, q$ .

The magnitude of this charge is $q = 7.35\; {\rm \mu C}$ . Apply the unit conversion $1\; {\rm \mu C} = 10^{-6}\; {\rm C}$ :

$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

$\begin{aligned}F &= E\, q \\ &= 5.25 \times 10^{5}\; {\rm N \cdot C^{-1}} \times (-7.35\times 10^{-6}\; {\rm C}) \\ &\approx 3.86\; {\rm N}\end{aligned}$ .

Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.

4 0

2 years ago