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3 years ago

The weight of a box is found to be 30 N. What is the approximate mass of the box?

Physics

2 answers:

svlad2 [7]3 years ago

6 0

Answer:

3kg

Explanation:

3kg is the appropriate mass

gtnhenbr [62]3 years ago

5 0

Answer:

3kg

Explanation:

just took the test

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Para que exista una fuerza se necesita cuando menos…

kozerog [31]

Answer:

Explanation:

No entiendo lo que quieres decir con eso

8 0

3 years ago

Which term describes the transfer of thermal energy from one object to another because of a difference in temperature? 1.energy

gregori [183]

The answer is 3.heat

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2 years ago

When a bus is moving (especially with a high speed) on the road suddenly stops or suddenly changes its direction, the luggage on

Masja [62]

Answer:

First law of motion

Explanation:

I say this because this example shows how an object is staying persistent unless it's compled to change

(not sure)

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2 years ago

The absolute pressure, in kilopascals, a depth 10m below sea level is most nearly?

saul85 [17]

Answer:

option A

Explanation:

given,

depth of the sea level = 10 m

g = 10 m/s²

Pressure underwater = ?

we know,

P = ρ g h

where ρ is the density of water which is equal to 1000 kg/m³

h is the depth of sea level

P = ρ g h

P = 1000 x 10 x 10

P = 100000 Pa

P = 100 kPa

Hence, the correct answer is option A

8 0

3 years ago

A dog leaps horizontally off a 70 m cliff with a speed of 6 m/s, how far from the base will the dog land?

iris [78.8K]

Answer:

$\Delta x=22.67786838m$

Explanation:

Let's use projectile motion equations. First of all we need to find the travel time. So we are going to use the next equation:

$y-y_0=v_o*sin(\theta)*t-\frac{1}{2}*t^2$ (1)

Where:

$y=Final\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$y_o=Initial\hspace{3}position\hspace{3}at\hspace{3}y-axis$

$v_o=initial\hspace{3}velocity$

$t=travel\hspace{3}time$

$g=gravity\hspace{3}constant$

$\theta=Initial\hspace{3}launch\hspace{3}angle$

In this case:

$\theta=0$

Because the dog jumps horizontally

Let's asume the gravity constant as:

$g=9.8$

$y=0$

Because when the dog reach the base the height is 0

$y_o=70$

$v_o=6$

Now let's replace the data in (1)

$y_o-\frac{1}{2} *(9.8)*t^2+70$

Isolating t:

$t=\pm\sqrt{\frac{2*70}{9.8} } =3.77964473$

Finally let's find the horizontal displacement using this equation:

$\Delta x=v_o*cos(\theta)*t$

Replacing the data:

$\Delta x=6*1*3.77964473=22.67786838m$

8 0

2 years ago