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aleksklad [387]

3 years ago

11

A car was traveling at a velocity of 8 m/s and then accelerated at a rate of 5 m/sr^2, reaching a velocity of 85 m/s. How far di

d the car travel at this acceleration rate?

1 answer:

juin [17]3 years ago

5 0

Answer:

716.1 m

Explanation:

A car was travelling at a velocity of 8m/s

It accelerated at 5m/s^2

It finally reached a velocity of 85m/s

The distance can be calculated by applying the fourth equation of motion

V^2= U^2 +2as

V= 85m/s

U= 8m/s

a= 5m/s^2

85^2= 8^2 + 2(5)(s)

7225= 64 + 10s

7225-64= 10s

7161=10s

s= 7161/10

= 716.1 m

Hence the car travelled at a distance of 716.1 m

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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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2 years ago

Which event is an example of melting?

qaws [65]

Answer:

welding metal

Explanation:

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2 years ago

The potential at location A is 382 V. A positively charged particle is released there from rest and arrives at location B with a

jarptica [38.1K]

Answer: 247.67 V

Explanation:

Given

Potential At A $V_a=382\ V$

Potential at $V_c=785\ V$

when particle starts from A it reaches with velocity $v_b$ at Point while when it starts from C it reaches at point B with velocity $2v_b$

Suppose m is the mass of Particle

Change in Kinetic Energy of particle moving under the Potential From A to B

$q\cdot \left ( V_a-V_b\right )=0.5m\cdot (v_b)^2----1$

Change in Kinetic Energy of particle moving under the Potential From C to B

$q\cdot \left ( V_c-V_b\right )=0.5m\cdot (2v_b)^2-----2$

Divide 1 and 2 we get

$\frac{V_a-V_b}{V_c-V_b}=\frac{v_b^2}{4v_b^2}$

on solving we get

$V_b=\frac{4}{3}\cdot V_a-\frac{1}{3}\cdot V_c$

$V_b=\frac{743}{3}=247.67\ V$

4 0

3 years ago

Giving brainliest!! if you answer correctly :) (30pts)

AnnZ [28]

Answer:

32000joule.

Explanation:

given

mass. (m)=160kg

speed (v)=20m/s

now

kinetic energy =1/2 (mv²)=1/2 ×{160×20²}=32000joule.

8 0

2 years ago

Read 2 more answers

The alarm at a fire station rings and a 87.5-kg fireman, starting from rest, slides down a pole to the floor below (a distance o

blsea [12.9K]

Answer:

$F_f=840N$

Explanation:

From the question we are told that

Weight of fireman $W_f= 87.5kg$

Pole distance $D=4.10m$

Final speed is $V_f 1.75m/s$

Generally the equation for velocity is mathematically represented as

$v^2 = v_0^2 + 2 a d$

Therefore Acceleration a

$a'= v^2 / 2 d$

$a'= 0.21m/s^2$

Generally the equation for Frictional force $F_f$ is mathematically given as

$F_f=m*a$

$F_f=m*(g-0.21)$

$F_f=87.5*(9.81-0.21)$

Therefore

$F_f=840N$

6 0

3 years ago