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Oxana [17]
2 years ago
12

An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass? show your work

Physics
1 answer:
amid [387]2 years ago
6 0

Answer:

Its  answer is 2.21 kg.

Explanation:

F =m × a

71 = m × 32

71 ÷ 32 = m

2.21 kg = m

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A

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A and B are isotopes of one another but the same element

C and D are isotopes of one another but the same element

However, A and B have a different proton count than C and D, indicating different elements because the proton count is equivalent to the atomic number.

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I will give Brainliest to whoever can answer!!!!
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The nodes and anti nodes would reverse roles.

Explanation:

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4 0
3 years ago
A dog, with a mass of 10.0 kg, is standing on a flatboat so that he is 22.5 m from the shore. He walks 7.8 m on the boat toward
marissa [1.9K]

Answer:16.096

Explanation:

Given

mass of dog\left ( m_d\right )=10kg

mass of boat\left ( m_b\right )=46kg

distance moved by dog relative to ground=x_d

distance moved by boat relative to ground=x_b

Distance moved by dog relative to boat=7.8m

There no net force on the system therefore centre of mass of system remains at its position

0=m_d\times x_d+m_b\dot x_b

0=10\times x_d+46\dot x_b

x_d=-4.6x_b

i.e. boat will move opposite to the direction of dog

Now

|x_d|+|x_b|=7.8

substitutingx_dvalue

5.6|x_b|=7.8

|x_b|=1.392m

|x_d|=6.4032m

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4 0
3 years ago
Which statement correctly describes the relationship between frequency and wavelength?
Reika [66]

Answer:

Answer:

Option B

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4 0
2 years ago
A physics professor did daredevil stunts in his spare time. His last stunt was an attempt to jump across a river on a motorcycle
tatiyna

Answer:

a) 17.8 m/s

b) 28.3 m

Explanation:

Given:

angle A = 53.0°

sinA = 0.8

cosA = 0.6

width of the river,d = 40.0 m,

the far bank was 15.0 m lower than the top of the ramp h = 15.0 m,

The river itself was 100 m below the ramp H = 100 m,

(a) find speed v

vertical displacement

-h= vsinA\times t-gt^2/2

putting values h=15 m, v=0.8

-15 = 0.8vt - 4.9t^2  ............. (1)

horizontal displacement d = vcosA×t = 0.6×v ×t

so v×t = d/0.6 = 40/0.6

plug it into (1) and get

-15 = 0.8\times40/0.6 - 4.9t^2

solving for t we get

t = 3.734 s

also, v = (40/0.6)/t = 40/(0.6×3.734) =  17.8 m/s

(b) If his speed was only half the value found in (a), where did he land?

v = 17.8/2 = 8.9 m/s

vertical displacement = -H =v sinA t - gt^2/2

⇒ 4.9t^2 - 8.9\times0.8t - 100 = 0

t = 5.30 s

then

d =v×cosA×t = 8.9×0.6×5.30= 28.3 m

3 0
3 years ago
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