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3 years ago

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Which table is it I’ll mark brainlest

1 answer:

alexira [117]3 years ago

5 0

Answer:

we need the graph to answer the question.

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To overcome an object's inertia, it must be acted upon by __________. A. gravity B. energy C. force D. acceleration

astra-53 [7]

In order to overcome an object’s inertia (resistance to change), it must be acted upon by an unbalanced force, so the answer to the problem is letter C.

7 0

3 years ago

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Two boxes are at rest on a smooth, horizontal surface. The boxes are in contact with one another. If box 1 is pushed with a forc

Maslowich

Answer:

mass of box 1 = 2.20 kg

mass of box 2 = 5.93 kg

Explanation:

Let the mass of box 1 and box 2 is respectively

$m_1$ and $m_2$

so we will have

Force applied on box 1 then acceleration

$a = \frac{F}{m_1 + m_2}$

$1.50 = \frac{12.2}{m_1 + m_2}$

$m_1 + m_2 = 8.13$

Now we know that contact force between them in above case is given as

$F_n = m_2a$

$8.90 = m_2(1.5)$

$m_2 = 5.93 kg$

now we have

$m_1 = 2.20 kg$

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3 years ago

Joaquin has hockey practice at the same ice rink where Alton has curling practice. One day, they were both playing on the ice wi

joja [24]

the puck recoils in each case.

larger mass stone gives puck greater recoil, smaller stone, smaller recoil

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3 years ago

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A proton is propelled at 4×106 m/s perpendicular to a uniform magnetic field. 1) If it experiences a magnetic force of 4.8×10−13

tia_tia [17]

Answer:

B = 0.75 T

Explanation:

As we know that the force on a moving charge in magnetic field is given by the formula

$F = qvB$

here we have

$B = \frac{F}{qv}$

here we know that

$F = 4.8 \times 10^{-13} N$

$q = 1.6 \times 10^{-19} C$

$v = 4 \times 10^6 m/s$

now from above equation we have

$B = \frac{4.8 \times 10^{-13}}{(1.6 \times 10^{-19})(4 \times 10^6)}$

$B = 0.75 T$

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3 years ago

A cosmic ray (an electron or nucleus moving ar speeds close to the speed of light) travels across the Milky Way at a speed of 0.

Fiesta28 [93]

Answer:

Cosmic ray's frame of reference: 99,875 years

Stationary frame of reference: 501,891 years

Explanation:

First of all, we convert the distance from parsec into metres:

$d=30,000 pc =9.26\cdot 10^{20} m$

The speed of the cosmic ray is

$v=0.98 c$

where

$c=3.0 \cdot 10^8 m/s$ is the speed of light. Substituting,

$v=(0.98)(3.0\cdot 10^8)=2.94\cdot 10^8 m/s$

And so, the time taken to complete the journey in the cosmic's ray frame of reference (called proper time) is:

$T_0 = \frac{d}{v}=\frac{9.26\cdot 10^{20}}{2.94\cdot 10^8}=3.15\cdot 10^{12} s$

Converting into years,

$T_0 = \frac{3.15\cdot 10^{12}}{(365\cdot 24\cdot 60 \cdot 60}=99,875 years$

Instead, the time elapsed in the stationary frame of reference is given by Lorentz transformation:

$T=\frac{T_0}{\sqrt{1-(\frac{v}{c^2})^2}}$

And substituting v = 0.98c, we find:

$T=\frac{99,875}{\sqrt{1-(\frac{0.98c}{c})^2}}=501,891 years$

3 0

3 years ago