3 years ago

What is the pressure exerted on three floor by a 10.000N crate if the bottom of the crate has the dimensions of 1/2m by 4m

Physics

1 answer:

Fantom [35]3 years ago

6 0

Answer:

fine the area then devide force by area

Explanation:

10000/(0.5*4)= 5000 pa

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a swimmer can swim in still water at a speed of 9.50 m/s. he intends to swim directly across the river that has a downstream cur

Doss [256]

Refer to the diagram shown below.

Still-water speed = 9.5 m/s
River speed = 3.75 m/s down stream.

The velocity of the swimmer relative to the bank is the vector sum of his still-water speed and the speed of the river.

The velocity relative to the bank is
V = √(9.5² + 3.75²) = 10.21 m/s

The downstream angle is
θ = tan⁻¹ 3.75/9.5 = 21.5°

Answer: 10.2 m/s at 21.5° downstream.

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3 years ago

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Consider the plot below describing motion along a straight line with an initial position of 10 m. −4 −3 −2 −1 0 1 2 3 4 5 6 1 2

Ugo [173]

Answer:

$+1 m/s^2$

Explanation:

Acceleration is given by

$a=\frac{\Delta v}{\Delta t}$

where

$\Delta v$ is the change in velocity

$\Delta t$ is the time interval in which the change in velocity occurs

To find the acceleration at 1 second, we can take the data at t = 1 s and t = 2. We find:

$\Delta t = -3 -(-4) = 1 s$

$\Delta v = 2 - (1) = +1 m/s$

So, the acceleration is

$a=\frac{+1}{1}=+1 m/s^2$

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3 years ago

imagine the first Olympic Games conducted on the moon in a specially designed dome. state and explain which sports would produce

Anastaziya [24]

Rhythmic gymnastics, trampoline gymnastics, javelin, diving, volleyball, and more due to the lack of gravity on the moon.

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3 years ago

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anyanavicka [17]

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3 years ago

El conductor de un tren que circula a 20 m/s ve un obstáculo en la vía y frena con una aceleración de 2 m/s2 hasta parar ¿cuánto

AlladinOne [14]

Velocidad inicial = 20 m/s
velocidad final = 0 m/s
aceleracion = -2 m/s^2

aceleracion = (cambio de velocidad)/(cambio de tiempo)
(cambio de tiempo)= (cambio de velocidad)/aceleracion
tiempo = (-20 m/s)/(-2 m/s^2)
= 10 segundos

x = (x(inicial)) + (v(inicial))(tiempo) + 1/2(aceleracion)(tiempo)^2
x(inicial) = 0
x = (20 m/s)(10 s) + 1/2 (-2m/s^2)(10 s)^2
x = 200 m - 100 m
x = 100 m (el espacio recorrido en los dos segundos)

espero que esto te ayude! buena suerte!

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3 years ago