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Sholpan [36]
2 years ago
8

Calculate the force needed to accelerate a baseball with a mass of 0.45

Physics
1 answer:
RideAnS [48]2 years ago
8 0
  • Mass=0.45kg
  • Acceleration=175m/s^2

\\ \bull\tt\dashrightarrow F=ma

\\ \bull\tt\dashrightarrow F=0.45(175)

\\ \bull\tt\dashrightarrow F=78.75N

\\ \bull\tt\dashrightarrow F=78.8N

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In a Young's double-slit experiment the separation distance y between the second-order bright fringe and the central bright frin
Natasha2012 [34]

Answer:

y = 0.0233 m

Explanation:

In a Young's Double Slit Experiment the distance between two consecutive bright fringes is given by the formula:

Δx = λL/d

where,

Δx = distance between fringes

λ = wavelength of light

L = Distance between screen and slits

d = Slit Separation

Now, for initial case:

λ = 425 nm = 4.25  x 10⁻⁷ m

y = 2Δx = 0.0177 m => Δx = 8.85 x 10⁻³ m

Therefore,

8.85 x 10⁻³ m = (4.25 x 10⁻⁷ m)L/d

L/d = (8.85 x 10⁻³ m)/(4.25 x 10⁻⁷ m)

L/d = 2.08 x 10⁴

using this for λ = 560 nm = 5.6 x 10⁻⁷ m:

Δx = (5.6 x 10⁻⁷ m)(2.08 x 10⁴)

Δx = 0.0116 m

and,

y = 2Δx

y = (2)(0.0116 m)

<u>y = 0.0233 m</u>

3 0
3 years ago
A 4 kg rock is dropped from 5 m. There is no friction. What kind of energy does is have before? What kind of energy does it have
denpristay [2]
1) The total mechanical energy of the rock is:
E=U+K
where U is the gravitational potential energy and K the kinetic energy.

Initially, the kinetic energy is zero (because the rock starts from rest, so its speed is zero), and the total mechanical energy of the rock is just gravitational potential energy. This is equal to
E_i=U=mgh
where m=4 kg is the mass, g=9.81 m/s^2 is the gravitational acceleration and h=5 m is the height.
Putting the numbers in, we find the potential energy
U=mgh=(4 kg)(9.81 m/s^2)(5 m)=196.2 J

2) Just before hitting the ground, the potential energy U is zero (because now h=0), and all the potential energy of the rock converted into kinetic energy, which is equal to:
E_f=K= \frac{1}{2}mv^2
where v is the speed of the rock just before hitting the ground. Since the mechanical energy of the rock must be conserved, then the kinetic energy K before hitting the ground must be equal to the initial potential energy U of the rock:
K=U=196.2 J

3) For the work-energy theorem, the work W done by the gravitational force on the rock is equal to the variation of kinetic energy of the rock, which is:
W=196.2 J-0 J=196.2 J
6 0
3 years ago
Refer to science 10 (mirror mirror on the wall)how do the height and width of the object compare with the height and width of th
Irina-Kira [14]

Depends on what type of mirror that is. I am going to assume this is a plain mirror (from the phrase), which means the height and width of the object and image is exactly the same.

6 0
2 years ago
A test charge of -3x10^-7 C is located 7 cm to the right of a charge of -9x10^-6 C and 20 cm to the left of a charge of +10x10^-
musickatia [10]

Answer:

5.634 N rightwards

Explanation:

qo = - 3 x 10^-7 C

q1 = - 9 x 10^-6 C

q2 = 10 x 10^-6 C

r1 = 7 cm = 0.07 m

r2 = 20 cm = 0.2 m

The force on test charge due to q1 is F1 which is acting towards right

According to the Coulomb's law

F_{1}=\frac{Kq_{1}q_{0}}{r_{1}^{2}}

F1 = (9 x 10^9 x 9 x 10^-6 x 3 x 10^-7) / (0.07 x 0.07)

F1 = 4.959 N rightwards

The force on test charge due to q2 is F1 which is acting towards right

According to the Coulomb's law

F_{2}=\frac{Kq_{2}q_{0}}{r_{2}^{2}}

F2 = (9 x 10^9 x 10 x 10^-6 x 3 x 10^-7) / (0.2 x 0.2)

F2 = 0.675 N rightwards

Net force on the test charge

F = F1 + F2 = 4.959 + 0.675 = 5.634 N rightwards

3 0
3 years ago
The moment of inertia of the empty turntable is 1.5 kg?m2. With a constant torque of 2.5 N?m, the turntableperson system takes 3
Tamiku [17]

6.0 \mathrm{kg} \mathrm{m}^{2} is the persons moment of inertia about an axis through her center of mass.

Answer: Option B

<u>Explanation:</u>

Given data are as follows:

moment of inertia of the empty turntable = 1.5

Torque = 2.5 N/m , and

           \text { Angular acceleration of the turntable }=\frac{\text { angular speed }}{\text { time }}=\frac{1}{3}

Let the persons moment of inertia about an axis through her center of mass= I

So, Now, from the formula of torque,

            \text { Torque }(\tau)=\text { Moment of inertia(I) } \times \text { Angular acceleration(a) }

            2.5=(1.5+I) \times \frac{1}{3}

So, from the above equation, we can measure the person’s moment of Inertia (I)

             2.5 \times 3=1.5+I

             I=7.5-1.5=6.0 \mathrm{kg} m^{2}

8 0
3 years ago
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