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3 years ago

Does anybody have the answered for this whole packet

Physics

2 answers:

Annette [7]3 years ago

8 0

No I don't have it I remember seeing that packet before tho

Monica [59]3 years ago

4 0

Sorry I've never seen this packet

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Two boxes on opposite ends of a massless board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on

rosijanka [135]

Answer:

b. 1.1 m

Explanation:

It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,

$s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)$

where,

s₁ = distance of fulcrum from left mass

s₂ = distance of fulcrum from right mass

In order to achieve balance, the torque due to both masses must be equal:

$T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}$

s₁ = 1.1 m

Hence, the correct option is:

4 0

3 years ago

A bat strikes a 0.145-kg baseball. Just before impact, the ball is traveling horizontally to the right at 60.0 m/s , and it leav

nata0808 [166]

Answer:

F = -307.4 N

Explanation:

It is given that,

Mass of the baseball, m = 0.145 kg

Initial speed of the baseball, u = 60 m/s

Final speed of the baseball, $v=65\ cos(30)=56.29\ m/s$

Time of contact, $t=1.75\ ms=1.75\times 10^{-3}\ s$

(a) It is assumed to find the horizontal component of average force. It is given by :

$F=m\dfrac{v-u}{t}$

$F=0.145\dfrac{56.29-60}{1.75\times 10^{-3}}$

F = -307.4 N

So, the horizontal component of average force is 307.4 N. Hence, this is the required solution.

8 0

3 years ago

The magnitude of Earth’s magnetic field is about 0.5 gauss near Earth’s surface. What’s the maximum possible magnetic force on a

vampirchik [111]

Answer:

$F = 1.5 \times 10^{-16} N$

this force is $1.68 \times 10^{13}$ times more than the gravitational force

Explanation:

Kinetic Energy of the electron is given as

$KE = 1 keV$

$KE = 1 \times 10^3 (1.6 \times 10^{-19}) J$

$KE = 1.6 \times 10^{-16} J$

now the speed of electron is given as

$KE = \frac{1}{2}mv^2$

now we have

$v = \sqrt{\frac{2 KE}{m}}$

$v = 1.87 \times 10^7 m/s$

now the maximum force due to magnetic field is given as

$F = qvB$

$F = (1.6\times 10^{-19})(1.87 \times 10^7)(0.5 \times 10^{-4})$

$F = 1.5 \times 10^{-16} N$

Now if this force is compared by the gravitational force on the electron then it is

$\frac{F}{F_g} = \frac{1.5 \times 10^{-16}}{9.1 \times 10^{-31} (9.8)}$

$\frac{F}{F_g} = 1.68 \times 10^{13}$

so this force is $1.68 \times 10^{13}$ times more than the gravitational force

4 0

3 years ago

Determine the kinetic energy of a 1000 kg roller coaster car that is moving with speed of 40.0 m/s

ololo11 [35]

Answer:

KE=800,000

Explanation:

The formula for kinetic energy is KE=1/2mv^2 or Kinetic Energy= 0.5*mass*velocity^2

so 1000 is the mass and 40 is the velocity

KE=0.5*1000*40^2

KE=0.5*1,000*1,600

KE=800,000 Joules

8 0

3 years ago

Give me the ans of this plsss

MissTica

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5 0

3 years ago