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NeX [460]
3 years ago
7

PLEASE SOLVE FAST!!! If the average American watches hours of TV every day , how many minutes will be spent in front of the TV b

y a person's 65th birthday? Solve using Dimensional Analysis.
Physics
1 answer:
masha68 [24]3 years ago
6 0

Answer:

5694000 min

Explanation:

Let's suppose the average American watches 4 hours of TV every day. First, we will calculate how many minutes they watch per day. We will use the conversion factor 1 h = 60 min.

(4 h/day) × (60 min/1 h) = 240 min/day

They watch 240 minutes of TV per day. Now, let's calculate how many minutes they watch per year. We will use the conversion factor 1 year = 365 day.

240 min/day × (365 day/year) = 87600 min/year

They watch 87600 min/year. Finally, let's calculate how many minutes they spend watching TV in 65 years.

87600 min/year × 65 year = 5694000 min

You might be interested in
a body of a mass 1kg suspended from a spring is found to stretch the spring by 10cm. (A) what is the spring constant? what is pe
Nikitich [7]

Answer:

(A) The spring constant is <u>98 N/m.</u>

(B) The period of oscillation is <u>0.635 s.</u>

Explanation:

Given:

Mass of the body is, m=1\ kg

Extension length of the spring is, x=10\ cm=0.1\ m

Now, let 'k' be the spring constant.

The force acting on the body is due to gravity only and is equal to its weight.

So, weight of the body is given as:

F_g=mg\\F_g=1\times 9.8\\F_g=9.8\ N

Now, we know that for a spring-mass system, the net force acting on the body is equal to the product of the spring constant and extension length. Therefore,

F_g=kx\\9.8=k(0.1)\\k=\frac{9.8}{0.1}=98\ N/m

Hence, the spring constant is 98 N/m.

(B)

Period of oscillation of a body of spring-mass system in SHM is given as:

T=2\pi\sqrt{\frac{m}{k}}

Plug in the given values and solve for period 'T'. This gives,

T=2\pi\sqrt{\frac{1}{98}}\\T=2\pi\times 0.101\\T=0.635\ s

Therefore, the period of oscillation is 0.635 s.

5 0
3 years ago
Yo-Yo man releases a yo-yo from rest and allows it to drop, as he keeps the top end of the string stationary. The mass of the yo
patriot [66]

To solve this problem it is necessary to use the conservation equations of both kinetic, rotational and potential energy.

By definition we know that

KE + KR = PE

Where,

KE =Kinetic Energy

KR = Rotational Kinetic Energy

PE = Potential Energy

In this way

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

Where,

m = mass

v= Velocity

I = Moment of Inertia

\omega = Angular velocity

g = Gravity

h = Height

We know as well that \omega = v/r for velocity (v) and Radius (r)

Therefore replacing we have

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

[tex]h= \frac{1}{2} \frac{v^2}{g} +\frac{1}{2} \frac{I}{mg}(\frac{v}{r})^2[/tex]

h= \frac{1}{2}v^2 ( \frac{1}{g} +\frac{I}{mg}\frac{1}{r^2} )

h= \frac{1}{2}0.75^2 ( \frac{1}{9.8} +\frac{2.9*10^{-5}}{(0.056)(9.8)}\frac{1}{(0.0064)^2} )

h = 0.3915m

Therefore the height must be 0.3915 for the yo-yo fall has a linear speed of 0.75m/s

6 0
3 years ago
Someone please helpp me out thanks !
elena-s [515]

Answer:

Silver.

Explanation:

To determine the identity of the metal, we need to calculate the density of the metal. This is illustrated below:

Mass of metal (m) = 18.15g

Length (L)= 1.2cm

Volume (V) = L³ = 1.2³ = 1.728cm³

Density =.?

The density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, it is expressed as:

Density = Mass /volume

With the above formula, we can obtain the density of the metal as follow:

Mass = 18.15g

Volume = 1.728cm³

Density =.?

Density = Mass /volume

Density = 18.15g/1.728cm³

Density of the metal = 10.50g/cm³

Comparing the density of metal obtained with the densities given in the table above, we can see that the density of the metal is the same with that of silver.

Therefore, the metal is silver.

8 0
3 years ago
If you placed a cylinder with a 0.25 m radius in the middle of a record player that spins at 4.7 rad/s, what is its angular mome
podryga [215]

Answer:

0.29 kg m² / s

Explanation:

Angular momentum is the product of moment of inertia and angular velocity.

L = I ω

For a solid cylinder, I = ½ m r².

L = ½ m r² ω

L = ½ (2.0 kg) (0.25 m)² (4.7 rad/s)

L = 0.29 kg m² / s

7 0
3 years ago
Consider a block of mass equal to 10kg sliding on an inclined plane of 30°, as shown in the figure below. The coefficient of kin
Dafna1 [17]

Answer:

(a) aₓ = 1.33 m/s² and aᵧ = -0.770 m/s²

(b) x = 0.665 m and y = 4.62 m

(c) 3.61 s

Explanation:

(a) There are two ways we can solve this.  The first way is to sum the forces in the x and y direction, then use the relation tan 30° = -aᵧ/aₓ, where aᵧ is the acceleration in the +y direction (up) and aₓ is the acceleration in the +x direction (right).

The second way is to sum the forces in the parallel and perpendicular directions to find the acceleration parallel to the incline, a.  Then, use the relations aᵧ = -a sin 30° and aₓ = a cos 30°.

Let's try the first method.  Sum of forces in the +y direction:

∑F = ma

N cos 30° + Nμ sin 30° − mg = maᵧ

N cos 30° + Nμ sin 30° − mg = -maₓ tan 30°

Sum of forces in the +x direction:

∑F = ma

N sin 30° − Nμ cos 30° = maₓ

Substituting:

N cos 30° + Nμ sin 30° − mg = -(N sin 30° − Nμ cos 30°) tan 30°

N cos 30° + Nμ sin 30° − mg = -N sin 30° tan 30° + Nμ sin 30°

N cos 30° − mg = -N sin 30° tan 30°

N (cos 30° + sin 30° tan 30°) = mg

N = mg / (cos 30° + sin 30° tan 30°)

N = (10 kg) (10 m/s²) / (cos 30° + sin 30° tan 30°)

N = 86.6 N

Now, solving for the accelerations:

N sin 30° − Nμ cos 30° = maₓ

aₓ = N (sin 30° − μ cos 30°) / m

aₓ = (86.6 N) (sin 30° − 0.4 cos 30°) / 10 kg

aₓ = 1.33 m/s²

N cos 30° + Nμ sin 30° − mg = maᵧ

aᵧ = N (cos 30° + μ sin 30°) / m − g

aᵧ = (86.6 N) (cos 30° + 0.4 sin 30°) / 10 kg − 10 m/s²

aᵧ = -0.770 m/s²

Now let's try the second method.

Sum of forces in the perpendicular direction:

∑F = ma

N − mg cos 30° = 0

N = mg cos 30°

Sum of forces in the parallel direction:

∑F = ma

mg sin 30° − Nμ = ma

mg sin 30° − mgμ cos 30° = ma

a = g (sin 30° − μ cos 30°)

a = (10 m/s²) (sin 30° − 0.4 cos 30°)

a = 1.536 m/s²

Solving for the accelerations:

aₓ = a cos 30°

aₓ = 1.33 m/s²

aᵧ = -a sin 30°

aᵧ = -0.770 m/s²

As you can see, the second method is faster and easier, but both methods will give you the same answer.

(b) In the x direction:

Given:

x₀ = 0 m

v₀ = 0 m/s

aₓ = 1.33 m/s²

t = 1 s

Find: x

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (1 s) + ½ (1.33 m/s²) (1 s)²

x = 0.665 m

In the y direction:

Given:

y₀ = 5 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

t = 1 s

Find: y

y = y₀ + v₀ t + ½ at²

y = 5 m + (0 m/s) (1 s) + ½ (-0.770 m/s²) (1 s)²

y = 4.62 m

(c) In the y direction:

Given:

y₀ = 5 m

y = 0 m

v₀ = 0 m/s

aᵧ = -0.770 m/s²

Find: t

y = y₀ + v₀ t + ½ at²

0 m = 5 m + (0 m/s) t + ½ (-0.770 m/s²) t²

t = 3.61 s

5 0
3 years ago
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