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2 years ago

A 500 kg motorcycle accelerates at a rate of 2 m/s .how much force was applied to the motorcycle?

Physics

2 answers:

Aleksandr [31]2 years ago

8 0

Answer:

by using formula F=ma which is m stand for mass a stand for acceleration. so 500kg × 2 ms^-2

Kamila [148]2 years ago

8 0

Hope this helped, click

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The intensity at a distance of 6.0 m from a source that is radiating equally in all directions is 6.0 × 10-10 w/m2 . what is the

satela [25.4K]

The intensity is defined as the ratio between the power emitted by the source and the area through which the power is calculated:
$I= \frac{P}{A}$ (1)
where
P is the power
A is the area

In our problem, the intensity is $I=6.0 \cdot 10^{-10} W/m^2$ . At a distance of r=6.0 m from the source, the area intercepted by the radiation (which propagates in all directions) is equal to the area of a sphere of radius r, so:
$A=4 \pi r^2 = 4 \pi (6.0 m)^2 = 452.2 m^2$

And so if we re-arrange (1) we find the power emitted by the source:
$P=IA = (6.0 \cdot 10^{-10}W/m^2)(452.2 m^2)=2.7 \cdot 10^{-7} W$

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2 years ago

How does the sun transfer energy to Earth?

aleksley [76]

Answer:

By electromagnetic waves.

Explanation:

The sun transfers heat to earth via electromagnetic waves in twomajor ways:

$Radiation-$ this is the transfer of energy by invisible electromagnetic ways.

$Convection-$ The radiant sun energy warms the atmosphere and becomes heat energy. This transfer of heat through movement of fluids or usually air is called convection.

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2 years ago

Read 2 more answers

A beam of microwaves with λ = 0.9 mm is incident upon a 9 cm slit. At a distance of 1.5 m from the slit, what is the approximate

liq [111]

Answer:

3 cm

Explanation:

According to the question,

$D=1.5 m$ .

$d=9 cm$ .

$\lambda =0.9 mm$ .

Now the approximate slit's image width is equal to width of central maxima.

And width of central maxima is twice the width from center to first maxima

So,

$y=2\frac{\lambda (D)}{d}$ .

Substitute all the variable in above equation.

$y=\frac{(2)0.9\times 10^{-2} m(1.5 m) }{0.09 m}$ .

$y=3 cm$ .

5 0

2 years ago

Sharks are generally negatively buoyant; the upward buoyant force is less than the weight force. This is one reason sharks tend

Tresset [83]

Answer:

8.67807 N

34.7123 N

Explanation:

m = Mass of shark = 92 kg

$\rho_{se}$ = Density of seawater = 1030 kg/m³

$\rho_{f}$ = Density of freshwater = 1000 kg/m³

$\rho_{sh}$ = Density of shark = 1040 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Net force on the fin is (seawater)

$F_n=mg-V_s\rho_{se}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{se}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1030\times 9.81\\\Rightarrow F_n=8.67807\ N$

The lift force required in seawater is 8.67807 N

Net force on the fin is (freshwater)

$F_n=mg-V_s\rho_{f}g\\\Rightarrow F_n=mg-\frac{m}{\rho_{sh}}\rho_{f}g\\\Rightarrow F_n=92\times 9.81-\frac{92}{1040}\times 1000\times 9.81\\\Rightarrow F_n=34.7123\ N$

The lift force required in a river is 34.7123 N

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2 years ago

The Lamborghini Huracan has an initial acceleration of 0.85g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al

SashulF [63]

Answer:

7.9 $\frac{m}{s^{2} }$

Explanation:

Take the fact that mass is inversely proportional to accelertation:

m ∝ a

Therefore m = a, but because we are finding the change in acceleration, we would set our problem up to look more like this:

$\frac{m_{1} }{m_{2} } = \frac{a_{2} }{a_{1} } \\$

Using algebra, we can rearrange our equation to find the final acceleration, $a_{2}$ :

$a_{2} = \frac{a_{1}*m_{1} }{m_{2} } \\$

Before plugging everything in, since you are being asked to find acceleration, you will want to convert 0.85g to m/s^2. To do this, multiply by g, which is equal to 9.8 m/s^2:

0.85g * 9.8 $\frac{m }{s^{2} }$ = 8.33 $\frac{m }{s^{2} }$

Plug everything in:

7.9 $\frac{m }{s^{2} }$ = $\frac{ 8.33\frac{m}{s^{2} }*1510kg }{1590kg}$

(1590kg the initial weight plus the weight of the added passenger)

8 0

1 year ago