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3 years ago

6

Tyler's favourite number is 8642, and he tries to fit it into everything he does. If he wishes to do that much work by climbing

stairs, how many stairs must he climb if each stair is an average of 18 cm tall and Tyler's mass is 64 kg?

1 answer:

Brums [2.3K]3 years ago

6 0

Answer:

2014

Explanation:

none

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Which of the following is equal to Velocity change per second

Mademuasel [1]

Acceleration

Explanation:

Acceleration is a physical quantity that expresses the change in the velocity of a body per unit of time.

Acceleration = $\frac{V - U }{T}$

V is the initial velocity

U is the final velocity

T is the time

It is has a unit of m/s²

Learn more:

Acceleration brainly.com/question/3820012

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2 years ago

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia

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Answer:

Una secadora de cabello tiene una resistencia de 10Ω al circular una corriente de 6 Amperes, si está conectado a una diferencia de potencial de 120 V, durante 18 minutos ¿Qué cantidad de calor produce?, expresado en calorías

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3 years ago

Professor blossom expects her students to be on time for class holds, them for the entire class period and has firm deadlines se

defon

Answer:

A dictatorship or tyranny.

Explanation:

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2 years ago

Does exists the friction force in space?If yes,tell an full example

VashaNatasha [74]

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3 years ago

A punter drops a 2.0 kg ball from rest vertically 1 meter down onto his foot. The ball leaves the foot with a speed of 18 m/s at

pav-90 [236]

Answer:

Explanation:

Given

mass of drop $m=2 kg$

height of fall $h=1 m$

ball leaves the foot with a speed of 18 m/s at an angle of $55^{\circ}$

Velocity of ball just before the collision with the floor

$u^=2gh$

$u=\sqrt{2gh}$

$u=\sqrt{2\times 9.8\times 1}=4.42 m/s$

Impulse delivered in Y direction

$J_y=m(v\sin (55)-(-u))$

$J_y=2(18\sin (55)+4.42)$

$J_y=38.32 kg-m/s$

Impulse in x direction

$J_x=m\times v\cos (55)$

$J_x=2\times 4.42\cos (55)=20.646$

$J_{net}=\sqrt{J_x^2+J_y^2}$

$J_{net}=\sqrt{(38.32)^2+(20.64)^2}$

$J_{net}=43.52 kg-m/s$

at an angle of $\tan \phi =\frac{J_y}{J_x}=\frac{38.32}{20.64}$

$\phi =tan^{-1}(1.856)$

$\phi =61.7^{\circ}$

7 0

2 years ago