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3 years ago

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An objec with a mass of 7 kg accelerates 5m/s2 when an unknown force is applied to it.what is the magnitude of the unknown force

?

1 answer:

OverLord2011 [107]3 years ago

4 0

Answer:

<h2>35 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 7 × 5

We have the final answer as

<h3>35 N</h3>

Hope this helps you

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A exaplme of a spoiler in flight <br>

Degger [83]

Answer:bad?

Explanation:

3 0

3 years ago

Read 2 more answers

A flywheel turns through 40 rev as it slows from an angular speed of 1.5 rad/s to a stop.

Lynna [10]

Answer:

-0.0047 rad/s²

335.103 seconds

99.18 seconds

Explanation:

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity = 1.5 ra/s

$\alpha$ = Angular acceleration

$\theta$ = Angle of rotation = 40 rev

t = Time taken

Equation of rotational motion

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{0^2-1.5^2}{2\times 2\pi \times 40}\\\Rightarrow \alpha=-0.0047\ rad/s^2$

Acceleration while slowing down is -0.0047 rad/s²

$t=\frac{\omega_f-\omega_i}{\alpha}\\\Rightarrow t=\frac{0-1.5}{-0.0047}\\\Rightarrow t=335.103\ s$

Time taken to slow down is 335.103 seconds

$\theta=\omega_it+\frac{1}{2}\alpha t^2\\\Rightarrow 20\times 2\pi=1.5\times t+\frac{1}{2}\times -0.0047\times t^2\\\Rightarrow 0.00235t^2-1.5t+125.66=0$

Solving the equation

$t=\frac{-\left(-1.5\right)+\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}, \frac{-\left(-1.5\right)-\sqrt{\left(-1.5\right)^2-4\cdot \:0.00235\cdot \:125.66}}{2\cdot \:0.00235}\\\Rightarrow t=539.11, 99.18\ s$

The time required for it to complete the first 20 is 99.18 seconds as 539.11>335.103

4 0

3 years ago

A positively charged wire with uniform charge density +λ lies along the x-axis and a negatively charged wire with uniform charge

Kisachek [45]

Answer:

$\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

Explanation:

The electric field created by an infinitely long wire can be found by Gauss' Law.

$\int \vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E2\pi r h = \frac{\lambda h}{\epsilon_0}\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 r} \^r$

For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.

$\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]$

5 0

3 years ago

Determine the average normal stress developed in rod AB if the load has a mass of 55 kg . The diameter of rod AB is 8 mm.

Fudgin [204]

Answer:

The normal stress is 10.7[MPa]

Explanation:

The normal stress can be calculated with the following equation:

$S_{norm} =\frac{F}{A} \\where:\\F= force [Newtons]\\A=area [m^2]\\S_{norm} = Normal stress [\frac{N}{m^{2} }] or [Pa]$

The area of the rod can be calculated using the equation:

$A=\frac{\pi }{4}*d^{2} \\d=8[mm]=0.008[m]\\A=\frac{\pi }{4}*(0.008)^{2} \\A=5.02*10^{-5} [m^{2} ]$

The force is the result of the mass multiplied by the gravity.

$F=55[kg]*9.81[m/s^{2} ] = 539.6[N]\\\\S_{norm} = 539.6/5.02*10^{-5} \\S_{norm} = 10.7*10^{6}[Pa] = 10.7[MPa]$

3 0

3 years ago

A conducting loop in the form of a circle is placed perpendicular to a magnetic field of 0. 50 t. If the area of the loop decrea

Kisachek [45]

The induced emf in the loop is -1500 μ V or - 0.0015 V .

According to the question

A conducting loop in the form of a circle is placed perpendicular to a magnetic field of 0. 50 t.

i.e

Magnetic field (B) = 0. 50 T

Area of circle or loop = $\pi r^{2}$

Now,

The area of the loop decreases at a rate of 3. 0 × 10⁻³ m/s

i.e

dA = 3. 0 × 10⁻³ meter²

dt = 1 sec

As per the formula of Induced e.m.f in the loop

emf is dependent on number of turns of coil, shape of the coil, strength of magnet and speed with which magnet is moved. Emf is independent of resistivity of wire of the coil.

$e=-B*\frac{dA}{dt}$

where A is the area of the loop.

Now ,

Substituting the values in the formula

$e=-B*\frac{dA}{dt}$

$e= - 0.50 *\frac{ 3 * 10^{-3}}{1}$

e = - 0.0015 V

OR

e = -1500 * 10⁻⁶ V

e = -1500 μ V

Negative just signifies emf will such be induced that current induced will oppose change in magnetic field though it

To know more about induced emf here:

brainly.com/question/16764848

#SPJ4

5 0

2 years ago