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3 years ago

14

In a football game, a 90 kg receiver leaps straight up in the air to catch the 0.42 kg ball the quarterback threw to him at a vi

gorous 21 m/s, catching the ball at the highest point in his jump. Right after catching the ball, how fast is the receiver moving?

1 answer:

irinina [24]3 years ago

8 0

To solve this problem it is necessary to apply the equations related to the conservation of momentum. Mathematically this can be expressed as

$m_1v_1+m_2v_2 = (m_1+m_2)v_f$

Where,

$m_{1,2}$ = Mass of each object

$v_{1,2}$ = Initial velocity of each object

$v_f$ = Final Velocity

Since the receiver's body is static for the initial velocity we have that the equation would become

$m_2v_2 = (m_1+m_2)v_f$

$(0.42)(21) = (90+0.42)v_f$

$v_f = 0.0975m/s$

Therefore the velocity right after catching the ball is 0.0975m/s

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Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

Answer: $m \frac{d}{dt}v_{(t)}$

Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

Because, if we derive velocity $v_{t}$ with respect to time $t$ we will have acceleration $a$ , hence:

$m \frac{d}{dt}v_{(t)}=m.a$

Where $m$ is the mass with units of kilograms ( $kg$ ) and $a$ with units of meter per square seconds $\frac{m}{s}^{2}$ , having as a result $kg\frac{m}{s}^{2}$

The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

$m\int x_{(t)} dt= m \frac{{(x_{(t)})}^{1+1}}{1+1}+C=m\frac{{(x_{(t)})}^{2}}{2}+C$ This result has units of $kgm^{2}$ and $C$ is a constant

$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

6 0

3 years ago

A vw beetle goes from 0 to 60 mi/h with an acceleration of 2.35 m/s^2.

Vlad1618 [11]

Part a.
u = 0, the initial velocity
v = 60 mi/h, the final velocity
a = 2.35 m/s², the acceleration.

Note that
1 m = 1609.34 m.
Therefore
v = (60 mi/h)*(1609.34 m/mi)*(1/3600 h/s) = 26.822 m/s
Use the formula
v = u + at
(26.822 m/s) = (2.35 m/s²)*(t s)
t = 26.822/2.35 = 11.4 s

Answer: 11.4 s

Part b.
We already determined that v = 60 mi/h = 26.822 m/s.
t = 0.6 s
Therefore
(26.822 m/s) = (a m/s²)*(0.6 s)
a = 26.822/0.6 = 44.7 m/s²

Answer: 44.7 m/s²

6 0

3 years ago

Express force in terms of base units

denis-greek [22]

Answer:

F = [MLT⁻²]

Explanation:

Force = ma

m (mass) = [M]

a (acceleration) = [LT⁻²]

F(force) = m x a = [MLT⁻²]

3 0

2 years ago

What's the difference between the speed and velocity of an object?

Artemon [7]

Speed is just how fast the object is moving, while velocity is the speed and <em>direction </em>of the object.

edit* Direction would be something such as North, East, etc.

4 0

3 years ago

Two spaceships are observed from earth to be approaching each other along a straight line. Ship A moves at 0.40c relative to the

nirvana33 [79]

Answer:

0.80 c

Explanation:

The computation of speed is shown below:-

Here, The speed of the captain ship A report for speed of the ship B which is

$S = \frac{S_A + S_B}{1 + \frac{(S_AS_B)}{c^2} }$

where

$S_A$ indicates the speed of the ship A

$S_B$ indicates the speed of the ship B

and

C indicates the velocity of life

now we will Substitute 0.40c for A and 0.60 for B in the equation which is

$S = \frac{0.40c + 0.60c}{1 + \frac{(0.40c)(0.60c)}{c^2} }$

after solving the above equation we will get

0.80 c

So, The correct answer is 0.80c

5 0

3 years ago