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3 years ago

14

How much force is needed to accelerate a 1-kilogram toy car at a rate of 2 meters per second per second?

2 answers:

Feliz [49]3 years ago

7 0

Answer:

2 Newtons

Explanation:

kozerog [31]3 years ago

5 0

F = ma, where m = mass in kg, a = acceleration in m/s², F = Force in Newton

F = 1 * 2

F = 2 N

Force needed is 2 Newtons.

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It is known that birds can detect the earth's magnetic field, but the mechanism of how they do this is not known. It has been su

san4es73 [151]

Answer:

$EMF = 5.01 \times 10^{-4} Volts$

Explanation:

Here we know that the EMF induced in this Field is given as

$EMF = vBL$

here B = perpendicular component of magnetic field

v = speed of the bird

L = length of the wings

now we have

$B = 5\times 10^{-5} sin40$

$v = 13 m/s$

$L = 1.2 m$

now we have

$EMF = (13)(3.21 \times 10^{-5})(1.2)$

$EMF = 5.01 \times 10^{-4} Volts$

8 0

3 years ago

A uniform rod of length 0.8 m and mass 1.4 kg, has two point masses at each end. The point mass on the left end has a mass 1.2 k

VladimirAG [237]

Answer:

Explanation:

1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m

5 0

2 years ago

Returning once again to our table top example of a horizontal mass on a low-friction surface with m = 0.254 kg and k = 10.0 N/m

Julli [10]

Explanation:

Given that,

Mass = 0.254 kg

Spring constant [tex[\omega_{0}= 10.0\ N/m[/tex]

Force = 0.5 N

y = 0.628

We need to calculate the A and d

Using formula of A and d

$A=\dfrac{\dfrac{F_{0}}{m}}{\sqrt{(\omega_{0}^2-\omega^{2})^2+y^2\omega^2}}$ .....(I)

$tan d=\dfrac{y\omega}{(\omega^2-\omega^2)}$ ....(II)

Put the value of $\omega=0.628\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-0.628)^2+0.628^2\times0.628^2}}$

$A=0.0198$

From equation (II)

$tan d=\dfrac{0.628\times0.628}{((10.0^2-0.628)^2)}$

$d=0.0023$

Put the value of $\omega=3.14\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-3.14)^2+0.628^2\times3.14^2}}$

$A=0.0203$

From equation (II)

$tan d=\dfrac{0.628\times3.14}{((10.0^2-3.14)^2)}$

$d=0.0120$

Put the value of $\omega=6.28\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-6.28)^2+0.628^2\times6.28^2}}$

$A=0.0209$

From equation (II)

$tan d=\dfrac{0.628\times6.28}{((10.0^2-6.28)^2)}$

$d=0.0257$

Put the value of $\omega=9.42\ rad/s$ in equation (I) and (II)

$A=\dfrac{\dfrac{0.5}{0.254}}{\sqrt{(10.0^2-9.42)^2+0.628^2\times9.42^2}}$

$A=0.0217$

From equation (II)

$tan d=\dfrac{0.628\times9.42}{((10.0^2-9.42)^2)}$

$d=0.0413$

Hence, This is the required solution.

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3 years ago

Write 4,007,500,000 in scientific notation.

Vika [28.1K]

The answer is 4.0075 x 10^9

7 0

2 years ago

In the figure below, this “double” nozzle discharges water (at 10°C, density= 1000 kg/m3) into the atmosphere at a rate of 0.50

dezoksy [38]

Answer:

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8 0

2 years ago