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3 years ago

What is fire department of the day

Physics

2 answers:

fredd [130]3 years ago

6 0

Answer:

International Firefighters' Day is observed on May 4. It was instituted after a proposal was emailed out across the world on January 4, 1999 due to the deaths of five firefighters in tragic circumstances in a bushfire in Australia.

Andrej [43]3 years ago

5 0

Fire Service Day
International Firefighters' Day (IFFD) is observed on May 4. It was instituted after a proposal was emailed out across the world on January 4, 1999 due to the deaths of five firefighters in tragic circumstances in a bushfire in Australia.

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A 35.9 g mass is attached to a horizontal spring with a spring constant of 18.4 N/m and released from rest with an amplitude of

lidiya [134]

Answer:

7.74m/s

Explanation:

Mass = 35.9g = 0.0359kg

A = 39.5cm = 0.395m

K = 18.4N/m

At equilibrium position, there's total conservation of energy.

Total energy = kinetic energy + potential energy

Total Energy = K.E + P.E

½KA² = ½mv² + ½kx²

½KA² = ½(mv² + kx²)

KA² = mv² + kx²

Collect like terms

KA² - Kx² = mv²

K(A² - x²) = mv²

V² = k/m (A² - x²)

V = √(K/m (A² - x²) )

note x = ½A

V = √(k/m (A² - (½A)²)

V = √(k/m (A² - A²/4))

Resolve the fraction between A.

V = √(¾. K/m. A² )

V = √(¾ * (18.4/0.0359)*(0.395)²)

V = √(0.75 * 512.53 * 0.156)

V = √(59.966)

V = 7.74m/s

8 0

3 years ago

Read 2 more answers

Bella makes the 2.5m distance to her food bowl in 9.1 seconds. What is her average velocity?

e-lub [12.9K]

Distance=2.5m
Time=9.1s

Average Velocity=Total Distance/Total Time

$\\ \sf\longmapsto \dfrac{2.5}{9.1}$

$\\ \sf\longmapsto 0.3m/s$

7 0

3 years ago

A hoop (I = MR2) of mass 3 kg and radius 1.1 m is rolling at a center-of-mass speed of 11 m/s. An external force does 842 J of w

Scilla [17]

Answer:

$v_f = 20 m/s$

Explanation:

Since the hoop is rolling on the floor so its total kinetic energy is given as

$KE = \frac{1}{2}mv^2 + \frac{1}{2} I\omega^2$

now for pure rolling condition we will have

$v = R\omega$

also we have

$I = mR^2$

now we will have

$KE = \frac{1}{2}mv^2 + \frac{1}{2}(mR^2)\frac{v^2}{R^2}$

$KE = mv^2$

now by work energy theorem we can say

$W = KE_f - KE_i$

$842 J = mv_f^2 - mv_i^2$

$842 = 3(v_f^2) - 3\times 11^2$

now solve for final speed

$v_f = 20 m/s$

3 0

3 years ago

For an object in uniform circular motion, what can you say about the directions of the velocity, acceleration, and net force vec

zavuch27 [327]

Answer: b) The velocity vector is perpendicular to the acceleration vector; the acceleration vector is parallel to the net force vector.

Explanation: A change in velocity creates an acceleration. As the object rotates through the circular path it is constantly changing direction, and hence accelerating, which causes a constant force to act upon the object. This Force acts towards the center of curvature, directly toward the axis of rotation in a direction parallel to the acceleration of the body along the path. Because the object is moving perpendicular to the force, the path followed by the object is a circular one. Hence the velocity of the object is perpendicular to the acceleration.

3 0

3 years ago

Larger animals use proportionately less energy than smaller animals; that is, it takes less energy per kg to power an elephant t

marin [14]

Answer:

Part a)

$Q = 952 cal/day$