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Svetach [21]
4 years ago
6

And ice cream truck is going 25 m/s to the east it accelerates to 40 m/s in the same direction over three seconds what is it acc

eleration
Physics
2 answers:
Andrej [43]4 years ago
5 0

Explanation:

It is given that,

Initial velocity of the truck, u = 25 m/s

Final speed of the truck, v = 40 m/s

Time taken, t = 3 s

We need to find the acceleration of the truck. It can be calculated using formula as :

a=\dfrac{v-u}{t}

a=\dfrac{40-25}{3}

a=5\ m/s^2

So, the acceleration of the truck is 5 meter per second square. Hence, this is the required solution.

Harrizon [31]4 years ago
3 0
V=Vo+at        (velocity=initial velocity plus acceleration multiplied by time)

40m/s=25m/s+a(3s)
-25      -25

15m/s=a(3s)    divide by 3s on both sides
5m/s^2=a
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3 years ago
A 400 hp engine in a 1,600 kg car applies maximum force for 2 seconds to accelerate the car onto the
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Answer:

I will assume that “maximum force” implies the constant application of power  P  = 400 hp (international) to accelerating the vehicle. The force will therefore vary with speed as the vehicle accelerates. I will also assume that all engine energy goes into accelerating the vehicle, rather than rotating elements like its wheels.

In this case the 400 hp (equivalent to 298,280 watts) is applied for time  t  = 2 seconds. Therefore the kinetic energy of the vehicle is increased by:

ΔKE=Pt=(298,280)(2)=596,560  joules.

The initial kinetic energy is:

KEinitial=12mv2

=(0.5)(1600)(82)=51,200  joules.

Therefore final kinetic energy is:

KEfinal=KEinitial+ΔKE

=51,200+596,560

=647,760  joules

Therefore final vehicle velocity can be found:

KEfinal=12mv2

v=2KEfinalm−−−−−−−−√

=(2)(647,760)1600−−−−−−−−−−−√

= 28.455 m/s

Explanation:

4 0
3 years ago
A box that is resting on an inclined plane has what forces acting on it
miskamm [114]

A box residing on an inclined plane has following forces:

In the X- direction

Mg sin (theta) = horizontal  component of weight of the box

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5 0
4 years ago
A string with a mass density of 3 * 10^-3 kg/m is under a tension of 380 N and is fixed at both ends. One of its resonance frequ
Delvig [45]

Answer:

(a) the fundamental frequency of this string is 65 Hz

(b) the harmonics of the given frequencies are third and fourth respectively.

(c) the length of the string is 2.74 m

Explanation:

Given;

mass density of the string, μ = 3 x 10⁻³ kg/m

tension of the string, T = 380 N

resonating frequencies, 195 Hz and 260 N

For the given resonant frequencies;

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } ---(1)\\\\260 = \frac{n+1}{2l} \sqrt{\frac{T}{\mu} } ---(2)\\\\divide \ (2) \ by (1)\\\\\frac{260}{195} = \frac{n+1 }{n} \\\\260n = 195(n+1)\\\\260 n = 195 n + 195\\\\260n - 195n = 195\\\\65n = 195\\\\n = \frac{195}{65} \\\\n = 3

(c) From any of the equations, solve for Length of the string (L);

195 = \frac{n}{2l} \sqrt{\frac{T}{\mu} } \\\\195 = \frac{3}{2l}\sqrt{\frac{380}{3\times 10^{-3}} } \\\\l = \frac{3}{2\times 195}\sqrt{\frac{380}{3\times 10^{-3}} }\\\\l = 2.74 \ m

(a) the fundamental frequency is calculated as;

f_o = \frac{1}{2l} \sqrt{\frac{T}{\mu} } \\\\f_o = \frac{1}{2\times 2.74} \sqrt{\frac{380}{3\times 10^{-3} } }\\\\f_o =  65 \ Hz

(b) harmonics of the given frequencies;

the first harmonic (n = 1) = f₀ = 65 Hz

the second harmonic (n = 2) = 2f₀ = 130 Hz

the third harmonic (n = 3) = 3f₀ = 195 Hz

the fourth harmonic (n = 4) = 4f₀ = 260 Hz

Thus, the harmonics of the given frequencies are third and fourth respectively.

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