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denis23 [38]
2 years ago
13

A 8.5 kg brick is dropped onto a 6.5 kg toy truck, which is moving across a level floor at 0.50 m/s. With what velocity do the t

ruck and brick continue to move, after the brick has landed on the truck?
Physics
1 answer:
Kipish [7]2 years ago
3 0

Answer:0.2167 m/s

Explanation:

Given

mass of brick M=8.5 kg

mass of toy truck m=6.5 kg

the velocity of truck u=0.5 m/s

Suppose v is the velocity after brick is landed on the truck

There is no external force acting so momentum is conserved

mu=(M+m)v

v=\dfrac{6.5}{15}\times 0.5\\v=0.2167\ m/s

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¿cual es la velocidad de un auto que recorre 4566 metros en 4 minutos? expresar en km\h
Svetlanka [38]

Explanation:

(4566 m / 4 min) × (1 km / 1000 m) × (60 min / h) = 68.49 km/h

3 0
3 years ago
Skylar travels 50 meters N and then goes 30 meters W before coming straight back south 20 meters. What distance did she travel?
lesya [120]

Answer:

100\ \text{meters}.

Explanation:

Distance Skylar traveled North is 50\ \text{meters}

Then she traveled 30\ \text{meters} Westward.

After which she traveled 20\ \text{meters} towards the South.

The total distance traveled would be the sum of the distances.

50+30+20=100\ \text{meters}

The distance traveled by Skylar was is 100\ \text{meters}.

4 0
2 years ago
Find the wavelength λ of the 80.0-khz wave emitted by the bat. express your answer in millimeters.
vfiekz [6]

Answer:

4.29 millimeters

Explanation:

Bats emit ultrasound waves: in air, ultrasound waves travel at a speed of

v=343 m/s

The frequency of the waves emitted by this bat is:

f=80.0 kHz = 80,000 Hz

Therefore we can find the wavelength of the wave emitted by the bat by using the relationship between speed, frequency and wavelength:

\lambda=\frac{v}{f}=\frac{343 m/s}{80,000 Hz}=4.29\cdot 10^{-3} m=4.29 mm

4 0
3 years ago
Suppose the acceleration of the particle moving gin a circle of radius "r" with the uniform speed "v" is proportional to some po
garik1379 [7]
So i say the power or the radius cortex in my equinox when i look at this V is the determination of V
5 0
2 years ago
Salmon often jump waterfalls to reach their breeding grounds. Starting downstream, 3.18 m away from a waterfall 0.294 m in heigh
Karolina [17]

Answer:

v = 7.65 m/s

t = 0.5882 s

Explanation:

We are told that the salmon started downstream, 3.18 m away from a waterfall.

Thus, range = 3.18 m

Since the horizontal velocity component is constant, then;

Range = vcosθ × t

Thus,

vcosθ × t = 3.18 - - - (eq 1)

We are told the salmon reached a height of 0.294 m

Thus, using distance equation;

s = v_y•t + ½gt²

g will be negative since motion is against gravity.

s = v_y•t - ½gt²

Thus;

0.294 = v_y•t - ½gt²

v_y = vsinθ

Thus;

0.294 = vtsinθ - ½gt² - - - (eq 2)

From eq(1), making v the subject, we have;

v = 3.18/tcosθ

Plugging into eq 2,we have;

0.294 = (3.18/tcosθ)tsinθ - ½gt²

0.295 = 3.18tanθ - ½gt²

We are given g = 9.81 m/s² and θ = 45°

0.295 = (3.18 × tan 45) - ½(9.81) × t²

0.295 = 3.18 - 4.905t²

3.18 - 0.295 = 4.905t²

4.905t² = 2.885

t = √2.885/4.905

t = 0.5882 s

Thus;

v = 3.18/(0.5882 × cos45)

v = 7.65 m/s

8 0
3 years ago
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