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Snowcat [4.5K]
2 years ago
5

Maria throws two stones from the top edge of a building with a speed of 20 m/s. She throws one straight down and the other strai

ght up. The first one hits the street in a time t1. How much later is it before the second stone hits
Physics
1 answer:
Georgia [21]2 years ago
7 0

Maria throws two stones from the top edge of a building at a speed of 20 m/s, The time is mathematically given as

t=4.5 sec.

<h3>How much later is it before the second stone hits?</h3>

Generally, the equation for speed  is mathematically given as

V = u+at

Therefore

0 = 22-9.8xt.

t = 22 /9.8 sec

t = 2.24489.

The same time will be taken from max^m height to point O

The extra time is taken

t=4.49 sec.

t=4.5 sec.

Read more about Time

brainly.com/question/4931057

#SPJ1

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yan [13]

Explanation:

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5 0
3 years ago
_ acceleration occurs when an object speeds up.<br> Answer<br> Positive
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_ acceleration occurs when an object speeds up.

Answer

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3 years ago
Read 2 more answers
A wire of Nichrome (a nickel-chromium-iron alloy commonly used in heating elements) is 1.3 m long and 1.6 mm2 in cross-sectional
fredd [130]

Answer: 1.3 *10^6 Ω*m

Explanation: In order to explain this problem we have to use the following expression for the resistence:

R=L/(σ*A) where L and A are the length and teh area for the wire, respectively. σ is the conductivity of teh Nichrome.

Then, from mteh OHM law we have V=R*I so R=V/I=2/3.2=0.625 Ω

Finally we have:

σ=L/(R*A)=1.3/(0.625*1.6*10^-6)=1.3*10^6 Ω*m

5 0
3 years ago
The bigclaw snapping shrimp shown in (Figure 1) is aptly named--it has one big claw that snaps shut with remarkable speed. The p
leva [86]

1) 1.86\cdot 10^6 rad/s^2

2) 2418 rad/s

3) 27000 m/s^2

4) 36.3 m/s

Explanation:

1)

The angular acceleration of an object in rotation is the rate of change of angular velocity.

It can be calculated using the following suvat equation for angular motion:

\theta=\omega_i t +\frac{1}{2}\alpha t^2

where:

\theta is the angular displacement

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

\theta=90^{\circ} = \frac{\pi}{2}rad is the angular displacement

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

Solving for \alpha, we find:

\alpha = \frac{2(\theta-\omega_i t)}{t^2}=\frac{2(\pi/2)-0}{0.0013}=1.86\cdot 10^6 rad/s^2

2)

For an object in accelerated rotational motion, the final angular speed can be found by using another suvat equation:

\omega_f = \omega_i + \alpha t

where

\omega_i is the initial angular velocity

t is the time

\alpha is the angular acceleration

In this problem we have:

t = 1.3 ms = 0.0013 s is the time elapsed

\omega_i = 0 is the initial angular velocity

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

Therefore, the final angular speed is:

\omega_f = 0 + (1.86\cdot 10^6)(0.0013)=2418 rad/s

3)

The tangential acceleration is related to the angular acceleration by the following formula:

a_t = \alpha r

where

a_t is the tangential acceleration

\alpha is the angular acceleration

r is the distance of the point from the centre of rotation

Here we want to find the tangential acceleration of the tip of the claw, so:

\alpha = 1.86\cdot 10^6 rad/s is the angular acceleration

r = 1.5 cm = 0.015 m is the distance of the tip of the claw from the axis of rotation

Substituting,

a_t=(1.86\cdot 10^6)(0.015)=27900 m/s^2

4)

Since the tip of the claw is moving by uniformly accelerated motion, we can find its final speed using the suvat equation:

v=u+at

where

u is the initial linear speed

a is the tangential acceleration

t is the time elapsed

Here we have:

a=27900 m/s^2 (tangential acceleration)

u = 0 m/s (it starts from rest)

t = 1.3 ms = 0.0013 s is the time elapsed

Substituting,

v=0+(27900)(0.0013)=36.3 m/s

5 0
3 years ago
What causes a compass needle to point to geographic north?
morpeh [17]
I believe it’s “C”
Hope that’s help you(:
8 0
3 years ago
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