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3 years ago

What is the accelration of a 6.4 kg bowling ball if a force of 12n is applied to ir

Physics

1 answer:

Zielflug [23.3K]3 years ago

3 0

What the function would look like would be like this:

F = ma
F = 12n / 6.4 kg
= 1.9m/s^2

Hopefully this helped. :)

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a net force of 219 N is exterted on a rock. the rock has an acceleration of 3m/s^2 due to this force. what is the mass of the ro

Sonbull [250]

Answer:

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{219}{3} \\$

We have the final answer as

Hope this helps you

6 0

3 years ago

I NEED HELP PLEASE, THANKS! :)

mrs_skeptik [129]

Answer:

1. Largest force: C; smallest force: B; 2. ratio = 9:1

Explanation:

The formula for the force exerted between two charges is

$F=K\dfrac{ q_{1}q_{2}}{r^{2}}$

where K is the Coulomb constant.

q₁ and q₂ are also identical and constant, so Kq₁q₂ is also constant.

For simplicity, let's combine Kq₁q₂ into a single constant, k.

Then, we can write

$F=\dfrac{k}{r^{2}}$

1. Net force on each particle

Let's

Call the distance between adjacent charges d.
Remember that like charges repel and unlike charges attract.

Define forces exerted to the right as positive and those to the left as negative.

(a) Force on A

$\begin{array}{rcl}F_{A} & = & F_{B} + F_{C} + F_{D}\\& = & -\dfrac{k}{d^{2}} - \dfrac{k}{(2d)^{2}} +\dfrac{k}{(3d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(-1 - \dfrac{1}{4} + \dfrac{1}{9} \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-36 - 9 + 4}{36} \right)\\\\& = & \mathbf{-\dfrac{41}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(b) Force on B

$\begin{array}{rcl}F_{B} & = & F_{A} + F_{C} + F_{D}\\& = & \dfrac{k}{d^{2}} - \dfrac{k}{d^{2}} + \dfrac{k}{(2d)^{2}}\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1}{4} \right)\\\\& = &\mathbf{\dfrac{1}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(C) Force on C

$\begin{array}{rcl}F_{C} & = & F_{A} + F_{B} + F_{D}\\& = & \dfrac{k}{(2d)^{2}} + \dfrac{k}{d^{2}} + \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( \dfrac{1}{4} +1 + 1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{1 + 4 + 4}{4} \right)\\\\& = & \mathbf{\dfrac{9}{4} \dfrac{k}{d^{2}}}\\\\\end{array}$

(d) Force on D

$\begin{array}{rcl}F_{D} & = & F_{A} + F_{B} + F_{C}\\& = & -\dfrac{k}{(3d)^{2}} - \dfrac{k}{(2d)^{2}} - \dfrac{k}{d^{2}}\\& = & \dfrac{k}{d^{2}}\left( -\dfrac{1}{9} - \dfrac{1}{4} -1 \right)\\\\& = & \dfrac{k}{d^{2}}\left(\dfrac{-4 - 9 -36}{36} \right)\\\\& = & \mathbf{-\dfrac{49}{36} \dfrac{k}{d^{2}}}\\\\\end{array}$

(e) Relative net forces

In comparing net forces, we are interested in their magnitude, not their direction (sign), so we use their absolute values.

$F_{A} : F_{B} : F_{C} : F_{D} = \dfrac{41}{36} : \dfrac{1}{4} : \dfrac{9}{4} : \dfrac{49}{36}\ = 41 : 9 : 81 : 49\\\\\text{C experiences the largest net force.}\\\text{B experiences the smallest net force.}\\$

2. Ratio of largest force to smallest

$\dfrac{ F_{C}}{ F_{B}} = \dfrac{81}{9} = \mathbf{9:1}\\\\\text{The ratio of the largest force to the smallest is $\large \boxed{\mathbf{9:1}}$}$

7 0

3 years ago

Consider a short time span just before and after the spark plug in a gasoline engine ignites the fuel-air mixture and releases 1

Tju [1.3M]

Answer:

Temperature after ignition=7883.205 K

Explanation:

The number of moles is,

n=PV/RT

=(1.18x10^6)(47.9x10^-6)/8.314(325)

= 0.0209 moles

a) In this process volume is constant

Q=U

=nCv.dT

dT= Q/nCv

=1970/(1.5x8.314)(0.0209)

= 7558.205 K

The final temperature is,

= 7558.205+325

= 7883.205 K

5 0

3 years ago

Read 2 more answers

When a 58g tennis ball is served, it accelerates from rest to a constant speed of 36 m/s. The impact with the racket gives the b

inysia [295]

We first calculate the acceleration on the ball using:
2as = v² - u²; u = 0 because ball is initially at rest
a = (36)²/(2 x 0.35)
a = 1850 m/s²
F = ma
F = 0.058 x 1850
= 107.3 Newtons

5 0

3 years ago

Incident rays parallel to the axis of a concave mirror reflect parallel to the axis.

coldgirl [10]

No they don't. Incident rays parallel to the axis of a concave mirror
reflect from the mirror's surface and converge at its focal point.

3 0

3 years ago