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2 years ago

A man driving a car traveling at 20 m/sec slams on the brakes and decelerates at 3.25

Physics

1 answer:

zhuklara [117]2 years ago

8 0

Recall that

${v_f}^2-{v_i}^2=2a\Delta x$

where $v_i$ and $v_f$ are initial and final velocities, respectively; $a$ is acceleration; and $\Delta x$ is the net displacement, or distance if the object is moving in a single direction.

The car has initial speed 20 m/s and acceleration -3.25 m/s². It comes to a stop, so it has 0 final speed. Then

0² - (20 m/s)² = 2 (-3.25 m/s²) ∆x

∆x = (20 m/s)² / (7.5 m/s²) ≈ 53.3 m

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A 50 kg pitcher throws a baseball with a mass of 0. 15 kg. If the ball is thrown with a positive velocity of 35 m/s and there is

dsp73

The velocity of the pitcher at the given mass is 0.1 m/s.

The given parameters:

Mass of the pitcher, m₁ = 50 kg
Mass of the baseball, m₂ = 0.15 kg
Velocity of the ball, u₂ = 35 m/s

Let the velocity of the pitcher = u₁

Apply the principle of conservation of linear momentum to determine the velocity of the pitcher as shown below;

m₁u₁ = m₂u₂

$u_1 = \frac{m_2 u_2}{m_1} \\\\u_1 = \frac{0.15 \times 35}{50} \\\\u_1 = 0.105 \ m/s\\\\u_1 \approx 0.1 \ m/s$

Thus, the velocity of the pitcher at the given mass is 0.1 m/s.

Learn more about conservation of linear momentum here: brainly.com/question/13589460

4 0

1 year ago

Calculate the electric field at the center of a square

pantera1 [17]

Answer:

$E_y=1175510.2\ N.C^{-1}$

The Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

Explanation:

Given:

side of a square, $a=52.5\ cm$

charge on one corner of the square, $q_1=+45\times 10^{-6}\ C$

charge on the remaining 3 corners of the square, $q_2=q_3=q_4=-27\times 10^{-6}\ C$

Distance of the center from each corners $=\frac{1}{2} \times diagonals$

$diagonal=\sqrt{52.5^2+52.5^2}$

$diagonal=74.25\cm=0.7425\ m$

∴Distance of center from corners, $b=0.3712\ m$

Now, electric field due to charges is given as:

$E=\frac{1}{4\pi\epsilon_0}\times \frac{q}{b^2}$

For charge $q_1$ we have the field lines emerging out of the charge since it is positively charged:

$E_1=9\times 10^9\times \frac{45\times 10^{-6}}{0.3712^2}$

$E_1=2938775.5\ N.C^{-1}$

Force by each of the charges at the remaining corners:

$E_2=E_3=E_4=9\times 10^9\times \frac{27\times 10^{-6}}{0.3712^2}$

$E_2=E_3=E_4=1763265.3\ N.C^{-1}$

Now, net electric field in the vertical direction:

$E_y=E_1-E_4$

$E_y=1175510.2\ N.C^{-1}$

Now, net electric field in the horizontal direction:

$E_y=E_2-E_3$

$E_y=0\ N.C^{-1}$

So the Magnitude of electric field is in the upward direction as shown directly towards the charge $q_1$ .

8 0

2 years ago

Suppose a wheel with a tire mounted on it is rotating at the constant rate of 2.59 times a second. A tack is stuck in the tire a

Vesna [10]

Answer:

Tangential speed=5.4 m/s

Radial acceleration= $88.6m/s^2$

Explanation:

We are given that

Angular speed=2.59 rev/s

We know that

1 revolution= $2\pi rad$

2.59 rev= $2\pi\times 2.59=5.18\pi=5.18\times 3.14=16.27 rad/s$

By using $\pi=3.14$

Angular velocity= $\omega=16.27rad/s$

Distance from axis=r=0.329 m

Tangential speed= $r\omega=16.27\times 0.329=5.4m/s$

Radial acceleration= $\frac{v^2}{r}$

Radial acceleration= $\frac{(5.4)^2}{0.329}=88.6m/s^2$

6 0

3 years ago

Does frequency affect wavelength?

Vinil7 [7]

Answer:

yeah

Explanation:

as wavelength increases frequency decreases and it goes the same for the opposite way

6 0

2 years ago

How can the motion of an object that is NOT moving change?

Semenov [28]

Only if a force acts upon it, it can move.

6 0

2 years ago