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inna [77]
3 years ago
7

An OSU linebacker of mass 110.0 kg sacks a UM quarterback of mass 85.0 kg. Just after they collide, they are momentarily stuck t

ogether, and both are moving at a speed of 2.60 m/s. If the quarterback was at rest just before he was sacked, how fast was the linebacker moving just before the collision
Physics
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

The initial speed of the linebacker just before the collision was 4.6 m/s.

Explanation:

Mass of OSU linebacker, m = 110 kg

Mass of UM quarterback, m' = 85 kg

Just after they collide, they are momentarily stuck together, the common speed is, V = 2.6 m/s

Initial speed of quarterback is 0 as it was at rest initially, u' = 0

Let u was the initial speed of the linebacker just before the collision. As they struck together, the momentum remains conserved. So,

mu+m'u'=(m+m')V\\\\mu=(m+m')V\\\\u=\dfrac{(m+m')V}{m}\\\\u=\dfrac{(110+85)\times 2.6}{110}\\\\u=4.6\ m/s

So, the initial speed of the linebacker just before the collision was 4.6 m/s.

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According to Newton’s Third Law, action and reaction are
Komok [63]
<span>B. equal and in opposite directions</span>
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3 years ago
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Una caja de 5.0kg de masa se acelera desde el reposo a través del piso mediante una fuerza a una tasa de 2.0 /s2 durante 7.0s en
Nady [450]

Responder:

<h2>490 julios </h2>

Explicación:

Se dice que el trabajo se realiza cuando una fuerza aplicada a un objeto hace que el objeto se mueva a través de una distancia. El trabajo realizado por un cuerpo se expresa mediante la fórmula;

Workdone = Fuerza * Distancia

Como Fuerza = masa * aceleración,

Workdone = masa * aceleración * distancia

Masa dada = 5.0kg, aceleración = 2.0m / s² d =?

Para obtener d, usaremos una de las leyes del movimiento,

d = ut + 1 / 2at²

u = 0 (ya que el cuerpo acelera desde el reposo) yt = 7.0s

d = 0 + 1/2 (2) (7) ²

d = 49m

Workdone = 5 * 2 * 49

Workdone = 490 Julios

4 0
3 years ago
a truck was traveling at 16.6 miles per second and accelerates at a rate of 2.0 meters per second squared how much time is requi
Ivan

A truck was traveling at 16.6 miles per second and accelerates at a rate of 2.0 meters per second squared then time is required for the truck to reach a speed of 25 miles per second is 6759 s.

Explanation:

Velocity is defined as the rate of change in displacement while acceleration is defined as the rate of change of velocity. Acceleration may be positive or negative. Acceleration is positive when the velocity of the object is increases and it is negative when velocity of the object is decreases. Negative acceleration is also called deceleration.

Mathematically

a = \frac{(v_{f} - v_{i})}{t}

Where a is the acceleration of the object, v_{f} is the final velocity of the object and  v_{i} is the initial velocity of the object. t is equal to time taken.

Given data:

v_{f} = 25 miles/s

v_{i} = 16.6 miles/s

a = 2.0 m/s²

t = ?

As velocities and acceleration given in different units, So we need to convert to obtain same units. Here we convert unit of acceleration  from m/s² to miles/s².

1 m/s² = 0.000621371192 miles/s²

2 m/s² = 0.00124274238 miles/s²

So,

a = 0.00124274238 miles/s²

Apply formula

a =\frac{v_{f} - v_{i}}{t}

t = \frac{(v_{f} - v_{i})}{a}

t = \frac{(25 - 16.6)}{0.00124274238}

t = 6759 s

Learn more about velocity and acceleration from

brainly.com/question/1192983

#learnwithBrainly

3 0
3 years ago
For Part A, Sebastian decided to use the copper cylinder. How would the magnitude of his q and ∆H compare if he were to redo Par
Vitek1552 [10]

The magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

The given parameters;

  • <em>initial temperature of metals, =  </em>T_m<em />
  • <em>initial temperature of water, = </em>T_i<em> </em>
  • <em>specific heat capacity of copper, </em>C_p<em> = 0.385 J/g.K</em>
  • <em>specific heat capacity of aluminum, </em>C_A = 0.9 J/g.K
  • <em>both metals have equal mass = m</em>

The quantity of heat transferred by each metal is calculated as follows;

Q = mcΔt

<em>For</em><em> copper metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_p = (m_wc_w + m_pc_p)(T_m - T_i)\\\\Q_p = (T_m - T_i)(m_wc_w ) + (T_m - T_i)(m_pc_p)\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m_p(T_m - T_i)\\\\m_p = m\\\\Q_p = (T_m - T_i)(m_wc_w ) + 0.385m(T_m - T_i)\\\\let \ (T_m - T_i)(m_wc_w )  = Q_i, \ \ \ and \ let \ (T_m- T_i) = \Delta t\\\\Q_p = Q_i + 0.385m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>copper metal</em>;

\Delta H = Q_p - Q_i\\\\\Delta H = (Q_i + 0.385m \Delta t) - Q_i\\\\\Delta H = 0.385 m \Delta t

<em>For </em><em>aluminum metal</em><em>, the quantity of heat transferred is calculated as</em>;

Q_A = (m_wc_w + m_Ac_A)(T_m - T_i)\\\\Q_A = (T_m -T_i)(m_wc_w) + (T_m -T_i) (m_Ac_A)\\\\let \ (T_m -T_i)(m_wc_w)  = Q_i, \ and \ let (T_m - T_i) = \Delta t\\\\Q_A = Q_i \ + \ m_Ac_A\Delta t\\\\m_A = m\\\\Q_A = Q_i \ + \ 0.9m\Delta t

<em>The </em><em>change</em><em> in </em><em>heat </em><em>energy for </em><em>aluminum metal </em><em>;</em>

\Delta H = Q_A - Q_i\\\\\Delta H = (Q_i + 0.9m\Delta t) - Q_i\\\\\Delta H = 0.9m\Delta t

Thus, we can conclude that the magnitudes of his q and ∆H for the copper trial would be lower than the aluminum trial.

Learn more here:brainly.com/question/15345295

6 0
3 years ago
A 0.0100-kilogram grape hangs from a vine 1.50 meters from the ground. What is the potential energy of the grape?
ludmilkaskok [199]
The potential energy is defined as Ep=m*g*h where m is the mass of the body, g=9.81 m/s² and h is the height of the body. In our case m=0.01 kg and h=1.5 m. So when we input the values into the equation:

Ep=0.01*9.81*1.5= 0.14715 J. 

So the potential energy of a grape is Ep=0.14715 J. 
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3 years ago
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