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2 years ago

11

How much work is done on a small car if a 3150 N force is exerted to move it 75.5 m to the side of the road

1 answer:

irinina [24]2 years ago

6 0

Answer:

Explanation:

Work = Force times displacement. Therefore,

W = 3150(75.5) so

W = 238000 N*m

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If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

vovikov84 [41]

Complete Question

The speed of a transverse wave on a string of length L and mass m under T is given by the formula

$v=\sqrt{\frac{T}{(m/l)}}$

If the maximum tension in the simulation is 10.0 N, what is the linear mass density (m/L) of the string

Answer:

$(m/l)=\frac{10}{V^2}$

Explanation:

From the question we are told that

Speed of a transverse wave given by

$v=\sqrt{\frac{T}{(m/l)}}$

Maximum Tension is $T=10.0N$

Generally making $(m/l)$ subject from the equation mathematically we have

$v=\sqrt{\frac{T}{(m/l)}}$

$v^2=\frac{T}{(m/l)}$

$(m/l)=\frac{T}{V^2}$

$(m/l)=\frac{10}{V^2}$

Therefore the Linear mass in terms of Velocity is given by

$(m/l)=\frac{10}{V^2}$

8 0

3 years ago

P6: An object of mass m sits on a spring of constant k in an elevator that is accelerating upwards with acceleration a. a) In te

tankabanditka [31]

Answer:

(a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

Explanation:

Given that,

Acceleration = a

mass = m

spring constant = k

(a). We need to calculate the spring compressed

Using balance equation

$kx-mg=ma$

$x=\dfrac{ma+mg}{k}$ ....(I)

The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). If the compression is 2.5 times larger than it is when the mass sits in a still elevator,

The compression is given by

$x=2.5\times x_{0}$

Here, acceleration is zero

So, $x=2.5\times\dfrac{mg}{k}$

We need to calculate the acceleration

Put the value of x in equation (I)

$2.5\times \dfrac{mg}{k}=\dfrac{ma+mg}{k}$

$2.5\times\dfrac{mg}{k}=\dfrac{m}{k}(a+g)$

$a=2.5g-g$

$a=1.5g$

Hence, (a). The spring compressed is $\dfrac{ma+mg}{k}$ .

(b). The acceleration is 1.5 g.

8 0

3 years ago

An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50

Setler79 [48]

Answer:

very hard others will answer it

Explanation:

hard

6 0

2 years ago

A 10n falling object encounters 4n of air resistance. what is the net force on the object?

Margaret [11]

It would be 6n down.

5 0

3 years ago

Three charged particles are positioned in the xy plane: a 50-nC charge at y = 6 m on the y axis, a –80-nC charge at x = –4 m on

leva [86]

Answer:48 V

Explanation:

Given

Three charged particle with charge

$q_1=50\ nC at y=6\ m$

$q_2=-80\ nC at x=-4\ m$

$q_3=70\ nC at y=-6\ m$

Electric Potential is given by

$V=\frac{kQ}{r}$

Distance of $q_1$ from $x=8\ m$

$d_1=\sqrt{6^2+8^2}$

$d_1=\sqrt{36+64}$

$d_1=10\ m$

similarly $d_2=8-(-4)$

$d_2=12\ m$

$d_3=\sqrt{(-6)^2+8^2}$

$d_3=\sqrt{36+64}$

$d_3=10\ m$

Potential at $x=8\ m$ is

$V_{net}=\frac{kq_1}{d_1}+\frac{kq_2}{d_2}+\frac{kq_3}{d_3}$

$V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]$

$V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}$

$V_{net}=9\times 5.33$

$V_{net}=47.97\approx 48\ V$

5 0

3 years ago