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Vedmedyk [2.9K]
1 year ago
8

Which statement accurately describes the motion of the object in the graph above over 10 seconds? Group of answer choices The ob

ject is at rest for 3 seconds and then decreases velocity to 1 cm/s for 3 seconds. The object is at rest for 4 seconds and then moves forward for 4 seconds at 1 cm/s. The object moves forward at 1 cm/s, stops, and then increases velocity to 2 cm/s. The object moves forward at 1 cm/s, stops, and then continues at the same velocity.

Physics
1 answer:
timurjin [86]1 year ago
7 0

Answer:

Object appears to move forward at 1 cm/sec, then the velocity drops to zero for 3 sec and then moves forward at 2 cm/sec     (11 - 3) / (10 - 6) = 2 cm/sec

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Step2247 [10]

Answer:

The decelerating force is 3\times 10^{- 11}\ N

Solution:

As per the question:

Frontal Area, A = 10\ m^{2}

Speed of the spaceship, v = 1\times 10^{6}\ m/s

Mass density of dust, \rho_{d} = 3\times 10^{- 18}\ kg/m^{3}

Now, to calculate the average decelerating force exerted by the particle:

Mass,\ m = \rho_{d}V                                (1)

Volume, V = A\times v\times t

Thus substituting the value of volume, V in eqn (1):

m = \rho_{d}(Avt)

where

A = Area

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t = time

m = \rho_{d}(A\times v\times t)                  (2)

Momentum,\ p = \rho_{d}(Avt)v = \rho_{d}Av^{2}t

From Newton's second law of motion:

F = \frac{dp}{dt}

Thus differentiating w.r.t time 't':

F_{avg} = \frac{d}{dt}(\rho_{d}Av^{2}t) = \rho_{d}Av^{2}

where

F_{avg} = average decelerating force of the particle

Now, substituting suitable values in the above eqn:

F_{avg} = 3\times 10^{- 18}\times 10\times 1\times 10^{6} = 3\times 10^{- 11}\ N

4 0
1 year ago
An airplane is delivering food to a small island. It flies 100 m above the ground at a speed of 150 m/s .
miss Akunina [59]

Answer:

The airplane should release the parcel 6.7*10^2 m before reaching the island

Explanation:

The height of the plane is y_0=100m, and its speed is v=150 m/s

When an object moves horizontally in free air (no friction), the equation for the y measured with respect to ground is

y=y_0 - \frac{gt^2}{2}    [1]

And the distance X is

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Being t the time elapsed since the release of the parcel

If we isolate t from the equation [1] and replace it in equation [2] we get

X = V . \sqrt{\frac{2y_0}{g}}

Using the given values:

x = 150 m/s  \sqrt{\frac{2\times 100m}{9.8 m/sec^2}}

x = 6.7*10^2 m

4 0
2 years ago
What is the advantage of the SI unit over CGS units?
MatroZZZ [7]

Answer:

The advantage of the SI unit over CGS unit are:

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1 year ago
The water level in a tank is 20 m above the ground. a hose is connected to the bottom of the tank, and the nozzle at the end of
Damm [24]

Answer:

P_(pump) = 98,000 Pa

Explanation:

We are given;

h2 = 30m

h1 = 20m

Density; ρ = 1000 kg/m³

First of all, we know that the sum of the pressures in the tank and the pump is equal to that of the Nozzle,

Thus, it can be expressed as;

P_(tank)+ P_(pump) = P_(nozzle)

Now, the pressure would be given by;

P = ρgh

So,

ρgh_1 + P_(pump) = ρgh_2

Thus,

P_(pump) = ρg(h_2 - h_1)

Plugging in the relevant values to obtain;

P_(pump) = 1000•9.8(30 - 20)

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REY [17]

Answer:

Explanation:

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