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3 years ago

Find the value of a x , the x-component of the object's acceleration.

1 answer:

3 0

To find the x component use the following formula, where Ф = theta = the angle 'a' makes with the x axis.

$a_x = a*cos(theta)$

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Bella makes the 2.5m distance to her food bowl in 9.1 seconds. What is her average velocity?

e-lub [12.9K]

Distance=2.5m
Time=9.1s

Average Velocity=Total Distance/Total Time

$\\ \sf\longmapsto \dfrac{2.5}{9.1}$

$\\ \sf\longmapsto 0.3m/s$

7 0

3 years ago

An electrical device in which a long wire is wound into a succession of closely spaced loops around a n insulating core is calle

Iteru [2.4K]

Answer:

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Explanation:

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3 years ago

What is the movement of a stationary object?

galben [10]

If it is stationary, its not moving. there is no movement

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3 years ago

An airplane in the process of taking off travels with a speed of 80 m/s at an angle of 15° above the horizontal. What is the gro

9966 [12]

Answer:

Option C

Explanation:

given,

velocity of airplane = 80 m/s

angle with the horizontal = 15°

speed of the ground= ?

when the plane is taking off the horizontal component of the velocity is v cosθ

so,

ground speed of the airplane is = $v\times cos\theta$

= $80 \times cos 15^0$

v = 77.27 m/s

horizontal velocity of the air plane comes out to be 77.27 m/s ≅ 77 m/s

so, the correct option is Option C

5 0

3 years ago

A charge of 50 µC is placed on the y axis at y = 3.0 cm and a 77-µC charge is placed on the x axis at x = 4.0 cm. If both charge

zheka24 [161]

Answer:

The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

Explanation:

Given that,

One Charge = 50 μC

Distance on y axis = 3.0 cm

Second charge = 77 μC

Distance on x axis = 4.0 cm

We need to calculate the force on electron due to q₁

Using formula of force

$F_{1}=\dfrac{kq_{1}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}$

$F_{1}=8\times10^{-11}j\ \ N$

We need to calculate the force on electron due to q₂

Using formula of force

$F_{2}=\dfrac{kq_{2}q}{r^2}$

Here, q = charge of electron

Put the value into the formula

$F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}$

$F_{2}=6.93\times10^{-11}i\ \ N$

We need to calculate the net force

Using formula of net force

$F=F_{1}+F_{2}$

Put the value into the formula

$F=8\times10^{-11}j+6.93\times10^{-11}i$

The magnitude of the net force

$F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}$

$F=1.058\times10^{-10}\ N$

We need to calculate the acceleration of an electron

Using formula of force

$F = ma$

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}$

$a=1.2\times10^{20}\ m/s^2$

Hence, The acceleration of an electron is $1.2\times10^{20}\ m/s^2$

3 0

3 years ago