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aivan3 [116]
2 years ago
4

When a water gun is fired while being held horizontally at a height of 1.00 m above ground level, the water travels a horizontal

distance of 5.00 m. a child, who is holding the same gun in a horizontal position, is also sliding down a 45.0° incline at a constant speed of 2.00 m/s.if the child fires the gun when it is 1.00 m above the ground and the water takes 0.329 s to reach the ground, how far will the water travel horizontally?

Physics
1 answer:
hram777 [196]2 years ago
8 0
Refer to the diagrams shown below.

Neglect air resistance.

Case A.
Let t =  the time of flight.
Then for horizontal travel,
t = (5 m)/(u m/s) = 5/u s

For vertical travel,
1.0 m = (1/2)*(9.8 m/s²)*(5/u s)²
5/u = √(1/4.9) = 0.4518
u = 11.0668 m/s

Case B.
There are two contributions to the horizontal velocity
(i) 11.0668 m/s
(ii) (2 m/s)*cos(45) = 1.4142 m/s
The total horizontal velocity is
11.0668 + 1.4142 = 12.481 m/s

Because the time of flight is 0.329 s, the horizontal distance traveled is
d = (12.481 m/s)*(0.329 s) = 4.106 m

Answer: The horizontal distance traveled is 4.11 m

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IRISSAK [1]

Answer:

maximum speed 56 km/h

Explanation:

To apply Newton's second law to this system we create a reference system with the horizontal x-axis and the Vertical y-axis. In this system, normal is the only force that we must decompose

       sin 10 = Nx / N

      cos 10 = Ny / N

      Ny = N cos 10

     Nx = N sin 10

Let's develop Newton's equations on each axis

X axis

We include the force of friction towards the center of the curve because the high-speed car has to get out of the curve

     Nx + fr = m a

     a = v2 / r

     fr = mu N

     N sin10 + mu N = m v² / r

     N (sin10 + mu) = m v² / r

Y Axis  

     Ny -W = 0

     N cos 10 = mg

Let's solve these two equations,

    (mg / cos 10) (sin 10 + mu) = m v² / r

    g (tan 10 + μ / cos 10) = v² / r

    v² = r g (tan 10 + μ / cos 10)

They ask us for the maximum speed

   v² = 30.0 9.8 (tan 10+ 0.65 / cos 10)

   v² = 294 (0.8364)  

   v = √(245.9)

   v = 15.68 m / s

Let's reduce this to km / h

   v = 15.68 m / s (1 km / 1000m) (3600s / 1h)

   v = 56.45 km / h

This is the maximum speed so you don't skid

7 0
2 years ago
A horizontal water jet strikes a stationary vertical plate at a rate of 5 kg/s with a velocity of 35 km/hr. Assume that the wate
Nina [5.8K]

Answer:

48.6 N

Explanation:

rate of mass per second, dm/dt = 5 kg/s

Velocity, v = 35 km/hr = 9.72 m/s

Force acting on the plate

F = v x dm/dt

F = 9.72 x 5 = 48.6 N

Thus, the force acting on the plate is 48.6 N.

3 0
2 years ago
A body covers a semicircle of radius 7cm in 5s .find its linear speed
choli [55]

Ok so we are given the radius of 7cm and time of 5 seconds.

From the data we got we can calculate speed, frequency, perimeter and area of the semicircle.

Let's start with perimeter.

We know that perimeter of circle is 2\pi r so the perimeter of semicircle is \dfrac{2\pi r}{2} or simply \pi r

So the perimeter is equal to:

\pi r=\pi\cdot7\approx\boxed{22cm}

So this is the length of a curve or let's say the distance.

Now let's look at the linear speed s=\dfrac{d}{t} where d is distance and t time.

We know the distance and we know the time.

So let's calculate it.

s=\dfrac{d}{t}=\dfrac{22}{5}=\boxed{4.4\dfrac{cm}{s}}

Hope this helps.

r3t40

8 0
2 years ago
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Answer:

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We have to find the average acceleration over that 3 s period.

We know that

Average acceleration,a=\frac{v-u}{t}{t}

Using the formula

Average acceleration,a=\frac{38-18}{3}ft/s^2

Average acceleration,a=\frac{20}{3}ft/s^2

Average acceleration,a=6.67ft/s^2

Hence, the average acceleration=6.67ft/s^2

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svlad2 [7]

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7 0
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