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3 years ago

8

Determine the weight of an average physical science textbook whose mass is 3.1 kilograms. The acceleration due to gravity is 9.8

m/s2.

1 answer:

Mrrafil [7]3 years ago

5 0

Given:

mass is 3.1 kilograms

The acceleration due to gravity is 9.8m/s2

Required:

Weight

Solution:

W = mg

W = (3.1 kilograms)( 9.8m/s2)

W = 30.38 Newtons

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According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt

scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength $(\lambda _1)=450 nm$

Second wavelength $(\lambda _2)=700 nm$

According wien's diplacement law

$\lambda T=constant$

where $\lambda =wavelength$

T=Temperature

Let $T_1 and T_2$ be the temperatures corresponding to $\lambda _1 & \lambda _2$ respectively.

$\lambda _1\times T_1=\lambda _2\times T_2$

$\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}$

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Thus object with $\lambda 450 nm$ is 1.55 times hotter than object with wavelength $\lambda =700 nm$

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Two 110 kg bumper cars are moving toward each other in opposite directions. Car A is moving at 8 m/s and Car Z at –10 m/s when t

salantis [7]

Explanation:

Mass of bumper cars, $m_1=m_2=110\ kg$

Initial speed of car A, $u_1=8\ m/s$

Initial speed of car Z, $u_2=-10\ m/s$

Final speed of car A after the collision, $v_1=-10\ m/s$

We need to find the velocity of car Z after the collision. Let it is equal to $v_2$ . Using the conservation of momentum as :

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

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$v_2=\dfrac{-1320}{110}\ m/s$

$v_2=-12\ m/s$

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How did people studying lunar eclipses learn that Earth is round?

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Answer:

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A speed skater moving across frictionless ice at 8.4 m/s hits a 5.7 m -wide patch of rough ice. She slows steadily, then continu

malfutka [58]

Answer:

Acceleration, $a=-2.48\ m/s^2$

Explanation:

Initial speed of the skater, u = 8.4 m/s

Final speed of the skater, v = 6.5 m/s

It hits a 5.7 m wide patch of rough ice, s = 5.7 m

We need to find the acceleration on the rough ice. The third equation of motion gives the relationship between the speed and the distance covered. Mathematically, it is given by :

$v^2-u^2=2as$

$a=\dfrac{v^2-u^2}{2s}$

$a=\dfrac{(6.5)^2-(8.4)^2}{2\times 5.7}$

$a=-2.48\ m/s^2$

So, the acceleration on the rough ice $-2.48\ m/s^2$ and negative sign shows deceleration.

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3 years ago