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2 years ago

PLEASE !! Help

Physics

1 answer:

n200080 [17]2 years ago

8 0

Answer:

scalar,magnitude

Explanation:

scalar is an example of length/distance

magnitude is the length of a vector

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A football player kicks a ball at a 30o angle from the ground with an initial velocity of 15 m/s. What is the final velocity of

navik [9.2K]

Given that,

Angle = 30°

Initial velocity = 15 m/s

We need to calculate the time of flight

Using formula of time of flight

$T=\dfrac{2u\sin\theta}{g}$

Where, u = initial velocity

g = acceleration due to gravity

Put the value into the formula

$T=\dfrac{2\times15\sin30}{9.8}$

$T=1.5\ sec$

We need to calculate the final velocity of the ball

Using equation of motion

$v=u+gt$

$v=15+9.8\times1.5$

$v=29.7\ m/s$

Hence, The final velocity of the ball is 29.7 m/s.

7 0

3 years ago

Elena (60.0 kg) and Madison (65.0 kg) are ice-skating at the Rockefeller ice rink in New Yok city. Their friend Tanner sees Elen

Bas_tet [7]

1. +72.0 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

Taking "to the right" as positive direction, for Elena we have

m = 60.0 kg is the mass

v = +1.20 m/s is the velocity

So, Elena's momentum is

$p_e=(60.0 kg)(+1.20 m/s)=+72.0 kg m/s$

2. -162.5 kg m/s

Here Madison is moving in the opposite direction of Elena (to the left), so her velocity is

v = -2.50 m/s

while her mass is

m = 65.0 kg

Therefore, her momentum is

$p_m= (65.0 kg)(-2.50 m/s)=-162.5 kg m/s$

3. -90.5 kg m/s

The total momentum of Elena and Madison is equal to the algebraic sum of their momenta; taking into account the correct signs, we have:

$p=p_e + p_m = +72.0 kg m/s - 162.5 kg m/s =-90.5 kg m/s$

4. 0.72 m/s to the left

We can find the final speed of Elena and Madison by using the law of conservation of momentum. In fact, the final momentum must be equal to the initial momentum (before the collision).

The initial momentum is the one calculated at the previous step:

$p_i = -90.5 kg m/s$

while the final momentum (after the collision) is given by

$p_f = (m_e + m_m) v$

where

$m_e$ is Elena's mass

$m_m$ is Madison's mass

v is their final velocity

According to the law of conservation of momentum,

$p_i = p_f\\p_i = (m_e + m_m) v$

So we can find v:

$v=\frac{p_i}{m_e + m_m}=\frac{-90.5 kg m/s}{60.0 kg+65.0 kg}=-0.72 m/s$

and the direction is to the left, since the sign is negative.

8 0

3 years ago

On the way to the moon, the Apollo astronauts reach a point where the Moon’s gravitational pull is stronger than that of Earth’s

Drupady [299]

Answer:

rm = 38280860.6[m]

Explanation:

We can solve this problem by using Newton's universal gravitation law.

In the attached image we can find a schematic of the locations of the Earth and the moon and that the sum of the distances re plus rm will be equal to the distance given as initial data in the problem rt = 3.84 × 108 m

$r_{e} = distance earth to the astronaut [m].\\r_{m} = distance moon to the astronaut [m]\\r_{t} = total distance = 3.84*10^8[m]$

Now the key to solving this problem is to establish a point of equalisation of both forces, i.e. the point where the Earth pulls the astronaut with the same force as the moon pulls the astronaut.

Mathematically this equals:

$F_{e} = F_{m}\\F_{e} =G*\frac{m_{e} *m_{a}}{r_{e}^{2} } \\$

$F_{m} =G*\frac{m_{m}*m_{a} }{r_{m} ^{2} } \\where:\\G = gravity constant = 6.67*10^{-11}[\frac{N*m^{2} }{kg^{2} } ] \\m_{e}= earth's mass = 5.98*10^{24}[kg]\\ m_{a}= astronaut mass = 100[kg]\\m_{m}= moon's mass = 7.36*10^{22}[kg]$

When we match these equations the masses cancel out as the universal gravitational constant

$G*\frac{m_{e} *m_{a} }{r_{e}^{2} } = G*\frac{m_{m} *m_{a} }{r_{m}^{2} }\\\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} }$

To solve this equation we have to replace the first equation of related with the distances.

$\frac{m_{e} }{r_{e}^{2} } = \frac{m_{m} }{r_{m}^{2} } \\\frac{5.98*10^{24} }{(3.84*10^{8}-r_{m} )^{2} } = \frac{7.36*10^{22} }{r_{m}^{2} }\\81.25*r_{m}^{2}=r_{m}^{2}-768*10^{6}* r_{m}+1.47*10^{17} \\80.25*r_{m}^{2}+768*10^{6}* r_{m}-1.47*10^{17} =0$

Now, we have a second-degree equation, the only way to solve it is by using the formula of the quadratic equation.

$r_{m1,2}=\frac{-b+- \sqrt{b^{2}-4*a*c } }{2*a}\\ where:\\a=80.25\\b=768*10^{6} \\c = -1.47*10^{17} \\replacing:\\r_{m1,2}=\frac{-768*10^{6}+- \sqrt{(768*10^{6})^{2}-4*80.25*(-1.47*10^{17}) } }{2*80.25}\\\\r_{m1}= 38280860.6[m] \\r_{m2}=-2.97*10^{17} [m]$

We work with positive value

rm = 38280860.6[m] = 38280.86[km]

6 0

3 years ago

HELP ME PLEASE!!!!

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