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3 years ago

Tameika makes a table about sensory organs

Physics

1 answer:

gayaneshka [121]3 years ago

4 0

Answer:

I think its d

Explanation:

I'm not sure I'm sorry if I'm wrong

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In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th

lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, $\theta=20.0$ °

Mass of package is, $m=2.50\ kg$

Initial speed of package is, $u=2.00\ m/s$

Final speed of the package at the bottom is, $v=0\ m/s$

Distance of travel along the incline is, $d=12.0\ m$

Acceleration due to gravity is, $g=9.8\ m/s^2$

Let the coefficient of kinetic friction be $\mu$ .

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

$a=\mu g\cos\theta-g\sin\theta$ ----------------- 1

Now, using equation of motion, we have:

$v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)$

Solving for 'a', we get:

$-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2$

Now, plug in the value of 'a' in equation (1). This gives,

$\mu g\cos\theta-g\sin\theta=\frac{1}{6}$ ( Neglecting negative sign)

Plug in all the given values and solve for $\mu$ . This gives,

$9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382$

Therefore, the coefficient of kinetic friction is 0.382.

5 0

3 years ago

Blood contains positive and negative ions and therefore is aconductor. A blood vessel, therefore, can be viewed as anelectrical

CaHeK987 [17]

Answer:

<h2>Magnetic field required for the given induced EMF is 1.41 T</h2>

Explanation:

Potential difference across the blood vessel is given as

$E = vBd$

here we know that the speed is given as

$v = 14.8 cm/s$

$d = 4.80 mm$

$E = 1 mV$

now we have

$1 \times 10^{-3} = (14.8 \times 10^{-2})B(4.80 \times 10^{-3})$

$B = 1.41 T$

Now volume flow rate of the blood is given as

$Q = Av$

$Q = \frac{\pi d^2v}{4}$

from above equation we have

$v = \frac{E}{Bd}$

Now we have

$Q = \frac{\pi d^2\frac{E}{Bd}}{4}$

$Q = \frac{\pi E d}{4B}$

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3 years ago

Your car is initially at rest when you hit that gas and the car begins to accelerate. The forward force of the car is 5630 N whi

Neporo4naja [7]

330...................

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2 years ago

Can waves move or transport matter? Why or why not?

OLga [1]

They can.
Surfers use waves to move themselves around.

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2 years ago

A 6.0-kilogram block, sliding to the east across a horizontal, frictionless surface with a momentum of 30.0 kilogram · meters pe

Lina20 [59]

The final speed of the block after the collision with the obstacle is $\boxed{3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}}$ .

Further Explanation:

Given:

The mass of the block is $6.0\,{\text{kg}}$ .

The initial momentum of the block is $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/ {\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}$ .

The impulse imparted by the obstacle is $10\,{\text{N}} \cdot {\text{s}}$ .

Concept:

The block is sliding towards east and the impulse imparted by the obstacle is towards the obstacle is towards west on the block. It means that the impulse exerted by the obstacle will reduce the momentum of the block.

According to the impulse momentum theorem, the rate of change of momentum of the body is equal to the impulse imparted to the body.

The expression for the impulse momentum theorem is.

${p_f} - p{ & _i} = I$ …… (1)

Substitute $30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}$ for ${p_i}$ and $- 10\,{\text{N}} \cdot {\text{s}}$ for $I$ in equation (1).

$\begin{aligned}{p_f} &= - 10\,{\text{N}} \cdot {\text{s}} + 30\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}} \\&= 20\,{{{\text{kg}} \cdot {\text{m}}} \mathord{\left/{\vphantom {{{\text{kg}} \cdot {\text{m}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}$

The final momentum of the block can be expressed as:

${p_f} = m{v_f}$ …… (2)

Substitute $20\text{kg}\;\text{m/s}$ for ${p_f}$ and $6.0\,{\text{kg}}$ for $m$ in equation (2).

$\begin{aligned}20 &= 6 \times {v_f} \\ {v_f}&= \frac{{20}}{6}\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&= 3.33\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}} \\ \end{aligned}$

Thus, the final speed of the block after the collision with the obstacle is $\boxed{3.33\;\text{m/s}}$ .

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Answer Details:

Grade: High School

Chapter: Impulse-momentum theorem

Subject: Physics

Keywords: Impulse, imparted, obstacle, speed, momentum, the obstacle, impulse-momentum theorem, frictionless surface, speed of block after collision.

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3 years ago

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