In the elements square on the periodic table, the number with the greatest numerical value the

(a) An elevator of mass m moving upward has two forces acting on it: the upward force of tension in the cable and the downward f

Katen [24]

Answer: T is greater

Explanation:

Since the elevator is moving against gravity more work will be done on the rope

T= m(g+a)

8 0

3 years ago

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Which of the following is characteristic of proficient catching?

uysha [10]

Answer:

The correct answer is option D i.e. A and C

Explanation:

The correct answer is option D i.e. A and C

for proficient catching player must

- learn to absorbed the ball force

- moves the hang according to ball direction to hold the ball

- to catch ball at high height move the finger at higher position

- to catch ball at low height move the finger at lower position

5 0

3 years ago

Calculate the linear acceleration (in m/s2) of a car, the 0.310 m radius tires of which have an angular acceleration of 15.0 rad

love history [14]

Answer:

a) The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ , b) The tires did 7.46 revolutions in 2.50 seconds from rest.

Explanation:

a) A tire experiments a general plane motion, which is the sum of rotation and translation. The linear acceleration experimented by the car corresponds to the linear acceleration at the center of the tire with respect to the point of contact between tire and ground, whose magnitude is described by the following formula measured in meters per square second:

$\| \vec a \| = \sqrt{a_{r}^{2} + a_{t}^{2}}$

Where:

$a_{r}$ - Magnitude of the radial acceleration, measured in meters per square second.

$a_{t}$ - Magnitude of the tangent acceleration, measured in meters per square second.

Let suppose that tire is moving on a horizontal ground, since radius of curvature is too big, then radial acceleration tends to be zero. So that:

$\| \vec a \| = a_{t}$

$\| \vec a \| = r \cdot \alpha$

Where:

$\alpha$ - Angular acceleration, measured in radians per square second.

$r$ - Radius of rotation (Radius of a tire), measured in meters.

Given that $\alpha = 15\,\frac{rad}{s^{2}}$ and $r = 0.31\,m$ . The linear acceleration experimented by the car is:

$\| \vec a \| = (0.31\,m)\cdot \left(15\,\frac{rad}{s^{2}} \right)$

$\| \vec a \| = 4.65\,\frac{m}{s^{2}}$

The linear acceleration of the car is $4.65\,\frac{m}{s^{2}}$ .

b) Assuming that angular acceleration is constant, the following kinematic equation is used:

$\theta = \theta_{o} + \omega_{o}\cdot t + \frac{1}{2}\cdot \alpha \cdot t^{2}$

Where:

$\theta$ - Final angular position, measured in radians.

$\theta_{o}$ - Initial angular position, measured in radians.

$\omega_{o}$ - Initial angular speed, measured in radians per second.

$\alpha$ - Angular acceleration, measured in radians per square second.

$t$ - Time, measured in seconds.

If $\theta_{o} = 0\,rad$ , $\omega_{o} = 0\,\frac{rad}{s}$ , $\alpha = 15\,\frac{rad}{s^{2}}$ , the final angular position is:

$\theta = 0\,rad + \left(0\,\frac{rad}{s}\right)\cdot (2.50\,s) + \frac{1}{2}\cdot \left(15\,\frac{rad}{s^{2}}\right)\cdot (2.50\,s)^{2}$

$\theta = 46.875\,rad$

Let convert this outcome into revolutions: (1 revolution is equal to 2π radians)

$\theta = 7.46\,rev$

The tires did 7.46 revolutions in 2.50 seconds from rest.

3 0

3 years ago

A 0.405 kg mass is attached to a spring with a force constant of 26.3 N/m and released from rest a distance of 3.31 cm from the

miv72 [106K]

Answer:

0.231 m/s

Explanation:

m = mass attached to the spring = 0.405 kg

k = spring constant of spring = 26.3 N/m

x₀ = initial position = 3.31 cm = 0.0331 m

x = final position = (0.5) x₀ = (0.5) (0.0331) = 0.01655 m

v₀ = initial speed = 0 m/s

v = final speed = ?

Using conservation of energy

Initial kinetic energy + initial spring energy = Final kinetic energy + final spring energy

(0.5) m v₀² + (0.5) k x₀² = (0.5) m v² + (0.5) k x²

m v₀² + k x₀² = m v² + k x²

(0.405) (0)² + (26.3) (0.0331)² = (0.405) v² + (26.3) (0.01655)²

v = 0.231 m/s

8 0

4 years ago

Pls help me l will make it brainlest

nikklg [1K]

Answer:

0.04 °C⁻¹

Explanation:

First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.

=============================================================

Finding Linear Expansivity :

⇒ α = Final length - Original length / (Original length × ΔT)

⇒ α = 9 - 4 / (4 × 70 - 20)

⇒ α = 5 / 5 × 50

⇒ α = 0.02

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Finding Area Expansivity :

⇒ Area Expansivity = 2 × Linear Expansivity

⇒ β = 2 × α

⇒ β = 2 × 0.02

⇒ β = 0.04 °C⁻¹

3 0

2 years ago

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