What is the key characteristic for double displacement

A baseball has a mass of 145 g. A bat exerts a force of 18,400 N on the ball. What is the acceleration of the ball?

amid [387]

The correct formula to use for the situation given above is: F = MA, where F is the applied force, M is the mass of the object and A is the acceleration.

From the details given in the question, we are told that:

F = 18, 400N

M = 145 g = 145 / 1000 = 0.145 kg

A = ?

From the equation F = MA

A = F / M

A = 18,400 / 0.145 = 126,896.55 = 1.27 *10^5.

Therefore, the correct option is C.

3 0

3 years ago

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Harry Potter is chasing his nemesis Draco Malfoy during a quidditch match. Initially, Harry is 35m behind Draco, who has just sp

Sophie [7]

Answer:

the acceleration of harry is equal to 66.126 m/s²

Explanation:

given,

harry is 35 m behind Draco

speed of Draco = 40 m/s

original speed of harry = 50 m/s

acceleration = ?

time taken by the Draco

t = $\dfrac{r}{u} =\dfrac{75}{40}$

t = 1.875 s

distance covered by Harry

d = 35 + 175 = 210 m

to calculate the acceleration of harry

$s = u t+ \dfrac{1}{2}at^2$

$210 = 50\times 1.875+ \dfrac{1}{2}\times a\times 1.875^2$

a × 3.516 × 0.5 = 116.25

a = 66.126 m/s²

hence, the acceleration of harry is equal to 66.126 m/s²

3 0

3 years ago

On earth which of the following are being investigated to contain fusion reactions

Vera_Pavlovna [14]

Answer:

A. Inertial Confinement and B. Magnetic Confinement

3 0

2 years ago

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how do I solve how much time does it take for a satellite moving 15000 miles per hour travel to travel once around the earth 249

iogann1982 [59]

Answer:

The time taken by the satellite to orbit earth at its surface, t = 1.66 hr

Explanation:

Given data,

The velocity of the satellite, v = 15000 miles/hr

The distance of travel, d = 24901 miles,

It is equal to the circumference of earth,

So the time taken by the satellite to orbit earth at its surface,

v = d/t

t = d/v

Substituting the given values in the above equation,

t = 24901 / 15000

= 1.66 hr

Hence, the time taken by the satellite to orbit earth at its surface, t = 1.66 hr

3 0

3 years ago

A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

7 0

2 years ago