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3 years ago

The stress force that pulls on the crust where two plates are moving apart is called ______.

Physics

2 answers:

malfutka [58]3 years ago

7 0

The stress force that pulls on the crust where two plates are moving apart is called<u> "tension".</u>

A tension refers to a force along the length of a medium, particularly a power conveyed by an adaptable medium, for example, a rope or link. "Tension" originates from a Latin word signifying "to extend." Not unintentionally, the adaptable ropes that convey muscle powers to different parts of the body are called tendons. Any adaptable connector, for example, a string, rope, chain, wire, or link, can apply pulls just parallel to its length; in this manner, a power conveyed by an adaptable connector is a pressure with course parallel to the connector. Understand that strain is a draw in a connector.

azamat3 years ago

5 0

Answer:

The stress force that pulls on the crust where two plates are moving apart is called tension.

Explanation:

A tension refers to a force along the length of a medium, particularly a power conveyed by an adaptable medium, for example, a rope or link.

The adaptable ropes that convey muscle powers to different parts of the body are called tendons.

Any adaptable connector, for example, a string, rope, chain, wire, or link, can apply pulls just parallel to its length; in this manner, a power conveyed by an adaptable connector is a pressure with course parallel to the connector.

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3 years ago

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The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee

kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2> $v_f = 35.3 m/s$ </h2>

Explanation:

Part a)

As we know that drag force is given as

$F = \frac{C_d \rho A v^2}{2}$

$C_d = 0.35$

$A = \frac{\pi d^2}{4}$

$A = \frac{\pi(0.074)^2}{4}$

$A = 4.3 \times 10^{-3} m^2$

$v = 40.2 m/s$

so we have

$F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}$

$F = 1.46 N$

So acceleration of the ball is

$a = \frac{F}{m}$

$a = \frac{1.46}{0.145}$

$a = 10.1 m/s^2$

Part B)

As per kinematics we know that

$v_f^2 - v_i^2 = 2 a d$

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4 years ago

A 1.20 kg water balloon will break if it experiences more than 530 N of force. Your 'friend' whips the water balloon toward you

soldi70 [24.7K]

Answer:

t = 0.029s

Explanation:

In order to calculate the interaction time at the moment of catching the ball, you take into account that the force exerted on an object is also given by the change, on time, of its linear momentum:

$F=\frac{\Delta p}{\Delta t}=m\frac{\Delta v}{\Delta t}$ (1)

m: mass of the water balloon = 1.20kg

Δv: change in the speed of the balloon = v2 - v1

v2: final speed = 0m/s (the balloon stops in my hands)

v1: initial speed = 13.0m/s

Δt: interaction time = ?

The water balloon brakes if the force is more than 530N. You solve the equation (1) for Δt and replace the values of the other parameters:

$|F|=|530N|= |m\frac{v_2-v_1}{\Delta t}|\\\\|530N|=| (1.20kg)\frac{0m/s-13.0m/s}{\Delta t}|\\\\\Delta t=0.029s$

The interaction time to avoid that the water balloon breaks is 0.029s

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