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FinnZ [79.3K]
3 years ago
6

Together, two students exert a force of 825 N in pushing a car a distance of 35 m.

Physics
1 answer:
andrey2020 [161]3 years ago
6 0

Answer:

28875 Joule

Explanation:

W = F * d

W = 825*35 = 28875 J

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A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t
Jlenok [28]

Answer:

14.36 N

Explanation:

T_{1} = Tension in string 1

T_{2} = Tension in string 2

m_b = mass of the bar = 2.7 kg

W_b = weight of the bar

weight of the bar is given as

W_b = m_{b} g = (2.7) (9.8) = 26.46N

m_m = mass of the bar = 1.35 kg

W_m = weight of the monkey

weight of the monkey is given as

W_m = m_{m} g = (1.35) (9.8) = 13.23N

Using equilibrium of torque about left end

W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33 N

Using equilibrium of force in vertical direction

T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36 N

7 0
3 years ago
Determine the 3 standing waves for a 4 m length of rope.
strojnjashka [21]

Harmonics, Loop and Harmonic number

Hope this helps :)

7 0
3 years ago
Tar is hard to pour because it doesn’t flow easily
IgorC [24]

tar is very thick and viscous making it difficult to pour when cold which is why it's heated to extremely high temperatures when blacktop is being poured

6 0
3 years ago
Read 2 more answers
Una persona de 80kg intenta bajar de peso subiendo una montaña para quemar el equilante a una rebanada de pastel de chocolate (7
SCORPION-xisa [38]

Answer:

Explanation:

An 80kg person tries to lose weight by climbing a mountain to burn the equivalent of a large slice of a rich chocolate cake (700kcl). How high should you climb?

Given that,

A person of mass

M = 80kg

The person is climbing a stair case and he burn an energy of 700 kcl

We know that 1kcl = 4184J

Then, 700kcl = 700 × 4184J

700kcl = 2.9288 × 10^6 J

We want to find height of stairs the person will climb when he Loss 700kcl. The energy he burns will be converted to the potential energy

P.E is calculated using

P.E = mgh

Where

m is mass in kg

g is gravitational constant g = 9.81m/s2

h is the height of stairs he climb in metre

Then,

P.E = energy burn

mgh = 700kcl

80 × 9.81 × h = 2.9288 × 10^6

h = (2.9288 × 10^6) / (80 × 9.81)

h = 3731.91m

So, the person will climb 3731.91m before he losses 700kcl

In espanyol

Dado que,

Una persona de masa

M = 80 kg

La persona está subiendo una escalera y quema una energía de 700 kcl.

Sabemos que 1kcl = 4184J

Entonces, 700kcl = 700 × 4184J

700kcl = 2.9288 × 10 ^ 6 J

Queremos encontrar la altura de las escaleras que la persona subirá cuando pierda 700kcl. La energía que quema se convertirá en energía potencial.

P.E se calcula utilizando

P.E = mgh

Dónde

m es masa en kg

g es constante gravitacional g = 9.81m / s2

h es la altura de las escaleras que sube en metros

Entonces,

P.E = quema de energía

mgh = 700kcl

80 × 9.81 × h = 2.9288 × 10 ^ 6

h = (2.9288 × 10 ^ 6) / (80 × 9.81)

h = 3731.91m

Entonces, la persona subirá 3731.91m antes de perder 700kcl

5 0
3 years ago
An automobile with a mass of 1240.00 kg has 3.99 m between
Vitek1552 [10]

Answer:

948.15248\ N

5134.04751\ N

Explanation:

F_f = Force on front wheels

F_r = Force on rear wheels

Distance between CG and rear wheel = 3.99-0.622 = 3.368 m

F_r=1240\times 9.81-F_f

As the forces are conserved we have

(1240\times 9.81-F_f)3.368=F_f\times 0.622\\\Rightarrow 12164.4-F_f=\dfrac{0.622F_f}{3.368}\\\Rightarrow 12164.4-F_f=0.18467F_f\\\Rightarrow F_f=\dfrac{12164.4}{1.18467}\\\Rightarrow F_f=10268.09503\ N

On rear wheels

F_r=1240\times 9.81-10268.09503\\\Rightarrow F_r=1896.30497\ N

The force on each rear whees is \dfrac{1896.30497}{2}=948.15248\ N

Force on each front wheel is \dfrac{10268.09503}{2}=5134.04751\ N

8 0
3 years ago
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