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soldi70 [24.7K]

3 years ago

14

While being thrown, a net force of 132 N acts on a baseball (mass = 140g) for a period of 4.5 x 10^-2 sec. what is the magnitude

of the change in momentum of the ball?

1 answer:

Vera_Pavlovna [14]3 years ago

5 0

Answer:

5.94 N·s

Explanation:

F = 132 N

t = 0.045 s

Impulse = Ft = (132 N)(0.045 s) = 5.94 N·s

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Blood cell : ____________ :: bacteria : _________________

DIA [1.3K]

Blood cell : Eukaryotic cell
and
Bacteria : Prokaryotic cell.

3 0

3 years ago

vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$

Then, plugging in the given values, we obtain:

$k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m$

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0

3 years ago

A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring

sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0

3 years ago

Two college teams, A and B, are playing a game of tug of war. Team A applies a force of 103 newtons, but the rope does not move

Kipish [7]

Given that the rope is not moving (acceleration is zero), by the second Law of Newton (F=m*a), the net force acting on the rope is zero.

Then, the force applied by the team B equals the force applied by the tema A: 103 N.

7 0

3 years ago

Read 2 more answers

The gage pressure in a liquid at a depth of 3 m is read to be 39 kPa. Determine the gage pressure in the same liquid at a depth

ioda

Answer: 117 kPa

Explanation:

For the liquid at depth 3 m, the gauge pressure is equal to = P₁=39 kPa

For the liquid at depth 9m, the gauge pressure is equal to= P₂

Now we are given the condition that the liquid is same. That must imply that the density must be same throughout the depth.

So, For finding gauge pressure we have formula P= ρ * g * h

Also gravity also remains same for both liquids

So taking ratio of their respective pressures we have

$\frac{P_{1} }{\\P_2}$ = $\frac{density * g * h_1}{density * g * h_2}$

So $\frac{39}{P_2}$ = $\frac{3}{9}$

Or P₂= 39 * 3 = 117 kPa

5 0

3 years ago