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3 years ago

B. Jerome plays middle linebacker for South's varsity football team. In a game against

Physics

1 answer:

weqwewe [10]3 years ago

4 0

Answer:

Option D

670 Kg.m/s

Explanation:

Initial momentum is given by mv=82*5.6=459.2 Kg.m/s (taking eastward as positive)

Final momentum is also mv but v being westward direction, we take it negative

Final momentum=82*-2.5= -205 Kg.m/s

Change in momentum=Final momentum-Initial momentum=-205-459.2=-664.2 Kg.m/s

Impulse=change in momentum=664.2 Kg.m/s rounded off as 670 Kg.m/s

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Two men pull on a 31-kg box with forces 8.3 N and 6.6 N in opposite directions.

kykrilka [37]

The resultant acceleration of the box is 0.055 m/s² in the direction of the greater force.

The given parameters:

Mass of the box, m = 31 kg
Two applied forces, = 8.3 N and 6.6 N

<h3>Resultant force on the box</h3>

The resultant force on the box is calculated as follows;

$F_{net} = F_1 - F_2\\\\F_{net} = 8.3 \ N - \ 6.6 \ N\\\\F_{net} = 1.7 \ N$

<h3>Resultant acceleration of the box</h3>

The resultant acceleration of the box is calculated as follows;

$F = ma\\\\a = \frac{F}{m} \\\\a = \frac{1.7}{31} \\\\a = 0.055 \ m/s^2$

Thus, the resultant acceleration of the box is 0.055 m/s² in the direction of the greater force.

Learn more about resultant acceleration here: brainly.com/question/26216029

4 0

2 years ago

Una tractomula se desplaza con rapidez de 69 km/h. Cuando el conductor ve una vaca atravesada enmedio de la carretera, acciona l

Anestetic [448]

Answer:

Los datos que tenemos:

Rapidez: 69km/h

Tiempo que tarda en frenar = 4s.

Distancia inicial entre la tracto-mula y la vaca = 25m

Ok, la ecuación de desaceleración es:

D = (sf - si)/t

sf = velocidad final = 0m/s

si = velocidad inicial = 69km/h

t = tiempo = 4s

D = -69km/h/4s

ok, 1h = 3600s

D = (-69km/s)*1/(4*3600s) = -0.0048 km/s^2

Entonces la ecuación de aceleración es:

a(t) = -0.0048 km/s^2

Para la velocidad, integramos sobre el tiempo

v(t) = (-0.0048 km/s^2)*t + v0

donde v0 es la velocidad inicial, en este caso v0 = 69km/3600s = 0.0191km/s

v(t) = (-0.0048 km/s^2)*t + 0.0191km/s

Para la posición volvemos a integrar sobre el tiempo, esta vez suponemos la posición inicial igual a cero.

p(t) = (1/2)*(-0.0048 km/s^2)*t^2 + 0.0191m/s*t

Ahora, si p(t=4s) < 25m, esto implica que la tracto-mula no impacto con la vaca.

p(4s) = (1/2)*(-0.0048 km/s^2)*(4s)^2 + 0.0191km/s*4s = 0.038km

y 1km = 1000m

0.038km = 0.038*1000m = 38m

Entonces si, atropello a la vaca.

4 0

4 years ago

The image illustrates that as the distance between two objects increases, the force of gravity ____________. A) decreases. B) in

emmainna [20.7K]

The image is missing (however it's not necessary to solve the problem).

The correct answer is A) decreases, because the gravitational force is inversely proportional to the square of the distance. In fact, the magnitude of the gravitational force between two object of mass M and m, at a distance d one from each other, is
$F=G \frac{Mm}{d^2}$
where G is the gravitational constant. As can be seen from the formula, if the distance d between the two object increases, the intensity of the force decreases.

7 0

3 years ago

Read 2 more answers

Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t

elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0

h = 1/2 gt²

$t = \sqrt{\frac{2h}{g} }$

Horizontal distance covered = t x Vx

= $\sqrt{\frac{2h}{g} } \times V_x$

From this expression also

Horizontal distance covered is proportional to $\sqrt{h}$ .

7 0

3 years ago

Air is cooled in a process with constant pressure of 150 kPa. Before the process begins, air has a specific volume of 0.062 m^3/

Mama L [17]

Answer:

The pressure is constant, and it is P = 150kpa.

the specific volumes are:

initial = 0.062 m^3/kg

final = 0.027 m^3/kg.

Then, the specific work can be written as:

$W = \int\limits^{vf}_{vi} {Pdv} \, = P(vf - vi) = 150kPa*(0.0027 - 0.062)m^3/kg = -5.25 kPa*m^3/kg.$

The fact that the work is negative, means that we need to apply work to the air in order to compress it.

Now, to write it in more common units we have that:

1 kPa*m^3 = 1000J.

-5.25 kPa*m^3/kg = -5250 J/kg.

7 0

3 years ago