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stepladder [879]
3 years ago
11

IF U ANSWER I WILL PUT U AS BRAINIEST ANSWER AND GIVE U A THANK U

Physics
2 answers:
Bingel [31]3 years ago
8 0
<span>
At the Earth's surface, warm air expands and rises, creating
what is known as an area of low pressure.

Cold air is dense and sinks to the surface to create what is
known as an area of high pressure.</span>


svlad2 [7]3 years ago
8 0

low pressure, high pressure


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5. In a local bar, a customer slides an empty beer mug on the counter for a refill. The bartender does not see the mug, which sl
pishuonlain [190]

Answer:

sorry about this question

5 0
3 years ago
if a man hits a golf ball (0.2 kg) which accelerates at a rate of 20 m/s squared. what amount of force acted on the ball?
kkurt [141]

F = ma

We have mass = 0.2kg

and acceleration = 20 m/s^2

So..

F = (0.2)(20)

F = 4 N

6 0
3 years ago
Please help
Snezhnost [94]

Answer:

Range = 22.61 m

Explanation:

We can use the formula for the Range in flat ground, given by:

Range=v_i^2\frac{sin(2\theta)}{g}

which for our case renders:

Range=15^2\frac{sin(80^o)}{9.8} \approx 22.61\,\,m

4 0
3 years ago
A proton that has a mass m and is moving at 270 m/s in the i hat direction undergoes a head-on elastic collision with a stationa
Nataly_w [17]

Answer:

V_p = 267.258 m/s

V_n = 38.375 m/s      

Explanation:

using the law of the conservation of the linear momentum:

P_i = P_f

where P_i is the inicial momemtum and P_f is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

P_i = m(270 m/s)

P_f = mV_P + M_nV_n

where m is the mass of the proton and V_p is the velocity of the proton after the collision, M_n is the mass of the nucleus and V_n is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = mV_p + 14mV_n

then, m is cancelated and we have:

270 = V_p + 14V_n

This is a elastic collision, so the kinetic energy K is conservated. Then:

K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2

and

Kf = \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

then,

\frac{1}{2}m(270)^2 =  \frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2

here we can cancel the m and get:

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

now, we have two equations and two incognites:

270 = V_p + 14V_n  (eq. 1)

\frac{1}{2}(270)^2 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2

in the second equation, we have:

36450 =  \frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2  (eq. 2)

from this last equation we solve for V_n as:

V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

and replace in the other equation as:

270 = V_p + 14\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }

so,

V_p = -267.258 m/s

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

V_n = \frac{270+267.258}{14}

V_n = 38.375 m/s  

3 0
3 years ago
Identify and explain which ball most likely had the greatest speed.
deff fn [24]
Y, bc the height of the bounce back is higher than x
4 0
3 years ago
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