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3 years ago

An NFL wide receiver prospect runs 40 m in 4.5 seconds. What is the average speed of the wide receiver (m/s)?

Physics

1 answer:

g100num [7]3 years ago

5 0

Answer:

8.89 m/s

Explanation:

rate = distance/time

40m/4.5 sec = 8.8888889 m/s

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Give Example of mental flexibility

Inessa [10]

Answer:

If you ring the doorbell and no one opens the door, you'll infer that no one is home rather than continuing to ring the doorbell to an empty house. Being able to understand this and look for another solution is another example of mental flexibility.

Explanation:

6 0

3 years ago

What is the answer??

kati45 [8]

Answer:

lol do you still need help?

Explanation:

8 0

3 years ago

Two hypothetical planets of masses m1 and m2 and radii r1 and r 2 , respectively, are nearly at rest when they are an infinite d

Leto [7]

Answer:

(a) $v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) Kinetic Energy of planet with mass m₁, is KE₁ = 1.068×10³² J

Kinetic Energy of planet with mass m₂, KE₂ = 2.6696×10³¹ J

Explanation:

Here we have when their distance is d apart

$F_{1} = F_{2} =G\frac{m_{1}m_{2}}{d^{2}}$

Energy is given by

$Energy \,of \,attraction = -G\frac{m_{1}m_{2}}{d}}+\frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$

Conservation of linear momentum gives

m₁·v₁ = m₂·v₂

From which

v₂ = m₁·v₁/m₂

At equilibrium, we have;

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2$ which gives

$2G{m_{1}m_{2}}= d m_{1} v^2_1+ dm_{2} (\frac{m_1}{m_2}v_1)^2= dv^2_1(m_1+(\frac{m_1}{m_2} )^2)$

multiplying both sides by m₂/m₁, we have

$2Gm^2_{2}}= dv^2_1 m_2+dm_1v^2_1 =dv^2_1( m_2+m_1)$

Such that v₁ = $\sqrt{\frac{2Gm^2_2}{d(m_1+m_2)} }$

$v_1 = m_2\sqrt{\frac{2G}{d(m_1+m_2)} }$

Similarly, with v₁ = m₂·v₂/m₁, we have

$G\frac{m_{1}m_{2}}{d}} = \frac{1}{2} m_{1} v^2_1+ \frac{1}{2} m_{2} v^2_2\Rightarrow 2G{m_{1}m_{2}}= dm_{1} (\frac{m_2}{m_1}v_1)^2 +d m_{2} v^2_2= dv^2_2(m_2+(\frac{m_2}{m_1} )^2)$

From which we have;

$2G{m^2_{1}}= dm_{2} v_2^2 +d m_{1} v^2_2$ and

$v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} }$

The relative velocity = v₁ + v₂ = $v_1+v_2=m_1\sqrt{\frac{2G}{d(m_1+m_2)} } + m_2\sqrt{\frac{2G}{d(m_1+m_2)} } = (m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

v₁ + v₂ = $(m_1+m_2)\sqrt{\frac{2G}{d(m_1+m_2)} }$

(b) The kinetic energy KE = $\frac{1}{2}mv^2$

$KE_1= \frac{1}{2} m_{1} v^2_1 \, \, \, KE_2= \frac{1}{2} m_{2} v^2_2$

Just before they collide, d = r₁ + r₂ = 3×10⁶+5×10⁶ = 8×10⁶ m

$v_1 = 8\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ = 10333.696 m/s

$v_2 = 2\times10^{24}\sqrt{\frac{2\times6.67408 \times 10^{-11}} {8\times10^6(2.00\times10^{24}+8.00\times10^{24})} }$ =2583.424 m/s

KE₁ = 0.5×2.0×10²⁴× 10333.696² = 1.068×10³² J

KE₂ = 0.5×8.0×10²⁴× 2583.424² = 2.6696×10³¹ J.

7 0

3 years ago

Part 2: Identify the independent, dependent, and constant variables

diamong [38]

ANSWER:

IV, Type of dish detergent. DV, height of foam. CV, type of container, amount of water in container, temperature of water, time the container is agitated.

Explanation:

Independent variable(IV)- what you change during the experiment.
dependent variable(DV)- what you're measuring during an experiment. The dependent variable is DEPENDENT because it's results DEPEND on the independent variable at play.
Constant variables(CV)- things that do not change in order to isolate the tested variables as much as possible.

4 0

3 years ago

A 2000 kg car experiences a constant braking force

Marat540 [252]

In that case, there are three possible scenarios:

-- If the braking force is less than the force delivered by the engine,
then the car will continue to accelerate, and the brakes will eventually
overheat and erupt in flame.

-- If the braking force is exactly equal to the force delivered by the engine,
then the car will continue moving at a constant speed, and the brakes will
eventually overheat and erupt in flame.

-- If the braking force is greater than the force delivered by the engine,
then the car will slow down and eventually stop. If it stops soon enough,
then the absorption of kinetic energy by the brakes will end before the
brakes overheat and erupt in flame. Even if the engine is still delivering
force, the brakes can be kept locked in order to keep the car stopped ...
They do not absorb and dissipate any energy when the car is motionless.

4 0

3 years ago