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4 years ago

8

A 120kg defensive lineman tackles a stationary quarterback at 6.5 m/s it takes .25 seconds for the lineman to completely stop wh

at is the magnitude of force exerted on the lineman during the collision

1 answer:

DIA [1.3K]4 years ago

8 0

Answer:

F = -3120 N

Explanation:

Lets write out everything we have:

Lineman:

m = 120kg

V1 = 6.5m/s

v2 = 0

t = 0.25

We should find acceleration, to move on to the next step;

a = v2-v1 / t

a = 0 - 6.5 / 0.25

a = -26m/s^2

Now we can use the Fnet formula and solve for F;

Fnet = ma

fnet = (120)(-26)

F = -3120 N

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Which of the following is an example of changing physical capital?

rosijanka [135]

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3 years ago

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4 years ago

Example 4.6 provides a nice example of the overlap between kinematics and dynamics. It is known that the plane accelerates from

kodGreya [7K]

Answer:

$ax = 2.60m/s^{2}$ , $t = 26.92s$

Explanation:

The acceleration of the plane can be determined by means of the kinematic equation that correspond to a Uniformly Accelerated Rectilinear Motion.

$(vx)f^{2} = (vx)i^{2} + 2ax \Lambda x$ (1)

Where $(vx)f^{2}$ is the final velocity, $(vx)i^{2}$ is the initial velocity, $ax$ is the acceleration and $\Lambda x$ is the distance traveled.

Equation (1) can be rewritten in terms of ax:

$(vx)f^{2} - (vx)i^{2} = 2ax \Lambda x$

$2ax \Lambda x = (vx)f^{2} - (vx)i^{2}$

$ax = \frac{(vx)f^{2} - (vx)i^{2}}{2 \Lambda x}$ (2)

Since the plane starts from rest, its initial velocity will be zero ( $(vx) = 0$ ):

Replacing the values given in equation 2, it is gotten:

$ax = \frac{(70m/s)^{2} - (0m/s)^{2}}{2(940m)}$

$ax = \frac{4900m/s}{2(940m)}$

$ax = \frac{4900m/s}{1880m}$

$ax = 2.60m/s^{2}$

So, The acceleration of the plane is $2.60m/s^{2}$

Now that the acceleration is known, the next equation can be used to find out the time:

$(vx)f = (vx)i + axt$ (3)

Rewritten equation (3) in terms of t:

$t = \frac{(vx)f - (vx)i}{ax}$

$t = \frac{70m/s - 0m/s}{2.60m/s^{2}}$

$t = 26.92s$

<u>Hence, the plane takes 26.92 seconds to reach its take-off speed.</u>

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3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

Using $1cm^2=10^{-4} m^2$

We know that

$R=\frac{\rho l}{A}$

In series

$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago

A drone is flying horizontally when it runs out of power and begins to free fall from 16 m. No drag. If it lands 40 m away (in t

marysya [2.9K]

Answer:

the horizontal velocity while it was falling is 22.1 m/s.

Explanation:

Given;

height of fall, h = 16 m

horizontal distance, x = 40 m

The time to travel 16 m is calculated as;

$t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 16}{9.8} } \\\\t = 1.81 \ s$

The horizontal velocity is calculated as;

$v_x = \frac{X}{t} \\\\v_x = \frac{40}{1.81} \\\\v_x = 22.1 \ m/s$

Therefore, the horizontal velocity while it was falling is 22.1 m/s.

7 0

3 years ago