When a pendulum is at the center position, what is true of the kinetic and potential energy?

A car with an initial speed of 4.30 m/s accelerates at a rate of 3 ms^2. What is the car's speed after 5 s?

Lorico [155]

Answer:

v = u + at

u = 4.30 m/s

a = 3 m/s^2

t = 5 s

v = 4.30 + (3)(5)

v = 4.30 + 15

v = 19.30 m/s

Explanation:

Hope this helps!

5 0

3 years ago

A 4 cm diameter "bobber" with a mass of 3 grams floats on a pond. A thin, light fishing line is tied to the bottom of the bobber

Tasya [4]

Answer:

Explanation:

Calculate the volume of the lead

$V=\frac{m}{d}\\\\=\frac{10g}{11.3g'cm^3}$

Now calculate the bouyant force acting on the lead

$F_L = Vpg$

$F_L=(\frac{10g}{11.3g/cm^3} )(1g/cm^3)(9.8m/s^2)\\\\=8.673\times 10^{-3}N$

This force will act in upward direction

Gravitational force on the lead due to its mass will act in downward direction

Hence the difference of this two force

$T=mg-F_L\\\\=(10\times10^{-3}kg(9.8m/s^2)-8.673\times 10^{-3}\\\\=8.933\times10^{-3}N$

If V is the volume submerged in the water then bouyant force on the bobber is

$F_B=V'pg$

Equate bouyant force with the tension and gravitational force

$F_B=T_mg\\\\V'pg=\frac{(8.933\times10^{-2}N)+mg}{pg} \\\\V'=\frac{(8.933\times10^{-2}N)+mg}{pg}$

Now Total volume of bobble is

$\frac{V'}{V^B} =\frac{\frac{(8.933\times10^{-2})+Mg}{pg} }{\frac{4}{3} \pi R^3 }\times100\\\\=\frac{\frac{(8.933\times10^{-2})+(3)(9.8)}{(1000)(9.8)} }{\frac{4}{3} \pi (4.0\times10^{-2})^3 }\times100\\\\$

= $\large\boxed{4.52 \%}$

7 0

3 years ago

A runner has a temperature of 40°c and is giving off heat at the rate of 50cal/s (a) What is the rate of heat loss in watts? (b)

Andrew [12]

Answer:

(a) 209 Watt

(b) 4482.8 seconds

Explanation:

(a) P = 50×4.18

Where P = rate of heat loss in watt

P = 209 Watt

Applying,

Q = cm(t₁-t₂)................ Equation 1

Where Q = amount of heat given off, c = specific heat capacity capacity of human, m = mass of the person, t₁ and t₂ = initial and final temperature.

From the question,

Given: m = 90 kg, t₁ = 40°C, t₂ = 37°C

Constant: c = 3470 J/kg.K

Substtut these values into equation 1

Q = 90×3470(40-37)

Q = 936900 J

But,

P = Q/t.............. Equation 2

Where t = time

t = Q/P............ Equation 3

Given: P = 209 Watt, Q = 936900

Substitute into equation 3

t = 936900/209

t = 4482.8 seconds

5 0

3 years ago

You are riding your bike to the mall. You travel the first mile in 10 minutes. The last mile takes you 15 minutes. This is an ex

Vikki [24]

This is an example of slowing down because you encountered heavy traffic or just got tired. Also, the numbers show acceleration.

6 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

4 years ago