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2 years ago

5

Ask a member of the family to help you.Do the following activities and identify the skill/skills being excited . use a separate

sheet of paper .

1 answer:

Archy [21]2 years ago

3 0

Answer:

Ok. Thanks.

I'll try it out.

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This version of Einstein’s equation is often used directly to find what value?

stepladder [879]

A. THE ENERGY THAT IS RELEASED IN A NUCLEAR REACTION.

M =MASS

C^2= SPEED OF LIGHT SQUARED

8 0

3 years ago

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You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride

marin [14]

a) The spring constant is 12,103 N/m

b) The mass of the trailer 2,678 kg

c) The frequency of oscillation is 0.478 Hz

d) The time taken for 10 oscillations is 20.9 s

Explanation:

a)

When the two children jumps on board of the trailer, the two springs compresses by a certain amount

$\Delta x = 6.17 cm = 0.0617 m$

Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:

$2mg = k'\Delta x$ (1)

where

m = 76.2 kg is the mass of each children

$g=9.8 m/s^2$ is the acceleration of gravity

$k'$ is the equivalent spring constant of the 2-spring system

For two springs in parallel each with constant k,

$k'=k+k=2k$

Substituting into (1) and solving for k, we find:

$2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m$

b)

The period of the oscillating system is given by

$T=2\pi \sqrt{\frac{m}{k'}}$

where

And for the system in the problem, we know that

T = 2.09 s is the period of oscillation

m is the mass of the trailer

$k'=2k=2(12,103)=24,206 N/m$ is the equivalent spring constant of the system

Solving the equation for m, we find the mass of the trailer:

$m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg$

c)

The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:

$f=\frac{1}{T}$

where

f is the frequency

T is the period

In this problem, we have

T = 2.09 s is the period

Therefore, the frequency of oscillation is

$f=\frac{1}{2.09}=0.478 Hz$

d)

The period of the system is

T = 2.09 s

And this time is the time it takes for the trailer to complete one oscillation.

In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.

Therefore, the time needed for 10 oscillations is:

$t=10T=10(2.09)=20.9 s$

#LearnwithBrainly

7 0

3 years ago

The launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m, where

lora16 [44]

Answer:

1.6 m

Explanation:

Given that the launch velocity of a toy car launcher is determined to be 5 m/s. If the car is to be launched from a height of 0.5 m.

The time for landing should be calculated by using the second equation of motion formula

h = Ut + 1/2gt^2

Let U = 0

0.5 = 1/2 × 9.8 × t^2

0.5 = 4.9t^2

t^2 = 0.5 / 4.9

t^2 = 0.102

t = 0.32 s

The target should be placed so that the toy car lands on it at:

Distance = 5 × 0.32

distance = 1.597 m

Distance = 1.6 m

Therefore, the target should be placed so that the toy car lands on it 1.6 metres away.

7 0

3 years ago

Learning Goal: To practice Problem-Solving Strategy 15.1 Mechanical Waves. Waves on a string are described by the following gene

kaheart [24]

Answer:

so the transverse displacement is 0.089963 m

Explanation:

Given data

equation y(x,t)=Acos(kx−ωt)

speed v = 9.00 m/s

amplitude A = 9.00 × 10^−2 m

wavelength λ = 0.480 m

to find out

the transverse displacement

solution

we know

v = angular frequency / wave number

and

wave number = 2 $\pi$ / λ = 2 $\pi$ / 0.480 = 13.0899 $m^{-2}$

angular frequency = v k

angular frequency = 9.00 × 13.0899

angular frequency = 117.8097 rad/sec = 118 rad/sec

so

equation y(x,t)=Acos(kx−ωt)

y(x,t)=9.00 × 10^−2 cos(13.0899 x−118t)

when x =0 and and t = 0

maximum y(x,t)= 9.00 × 10^−2 cos(13.0899 (0) − 118 (0))

maximum y(x,t)= 9.00 × 10^−2 m

and when x = x = 1.59 m and t = 0.150 s

y(x,t)=9.00 × 10^−2 cos(13.0899 (1.59) −118(0.150) )

y(x,t)=9.00 × 10^−2 × (0.99959)

y(x,t) = 0.089963 m

so the transverse displacement is 0.089963 m

5 0

3 years ago

Q)How is a conductor different from an insulator. A)A Conductor contains free electrons. B)A conductor contains free protons. C)

I am Lyosha [343]

A conductor contains electrons that are bound so weakly to their atoms
that they can be ripped away with a small voltage, and sent scurrying in
the direction of the higher potential.

5 0

3 years ago

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