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4 years ago

Different forces were applied to five balls, and each force was applied for the same amount of time. The data is in the table.

Physics

2 answers:

emmainna [20.7K]4 years ago

6 0

C. It doubles I hope this helps and good luck

Alja [10]4 years ago

6 0

The impulse provided to the ball is equal to the change in momentum of the ball.

According to Newton's second law, The rate of change of momentum is directly proportional to the force applied. Hence, the equation can be written as:

Δp = F × t

Δp = change in momentum = Impulse

F = Force

t = time for which the force is applied

Now. if the force is doubled and the time remains same.

F(new) = 2F

t = t

Δp (new) = 2F × t

Δp (new) = 2 Δp (old)

Hence, the impulse gets doubled

Option C is correct.

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Estimate the number of photons emitted per second from 1.0 cm2 of a person's skin if a typical emitted photon has a wavelength o

Diano4ka-milaya [45]

Answer:

2.63 x 10^18

Explanation:

A = 1 cm^2 = 1 x 10^-4 m^2

λ = 10,000 nm = 10,000 x 10^-9 m = 10^-5 m

T = 37 degree C = 37 + 273 = 310 k

Energy of each photon = h c / λ

where, h is the Plank's constant and c be the velocity of light

Energy of each photon = (6.63 x 10^-34 x 3 x 10^8) / 10^-5 = 1.989 x 10^-20 J

Energy radiated per unit time = σ A T^4

Where, σ is Stefan's constant

Energy radiated per unit time = 5.67 x 10^-8 x 10^-4 x 310^4 = 0.05236 J

Number of photons per second = Energy radiated per unit time / Energy of

each photon

Number of photons per second = 0.05236 / (1.989 x 10^-20) = 2.63 x 10^18

4 0

3 years ago

A force of 3600 N is exerted on a piston that has an area of 0.030 m2. What force is exerted on a second piston that has an area

Airida [17]

For Pascal's law, the pressure is transmitted with equal intensity to every part of the fluid:
$p_1 = p_2$
which becomes
$\frac{F_1}{A_1}= \frac{F_2}{A_2}$
where
$F_1=3600 N$ is the force on the first piston
$A_1=0.030 m^2$ is the area of the first piston
$F_2$ is the force on the second piston
$A_2=0.015 m^2$ is the area of the second piston

If we rearrange the equation and we use these data, we can find the intensity of the force on the second piston:
$F_2=F_1 \frac{A_2}{A_1}=(3600 N) \frac{0.015 m^2}{0.030 m^2}= 1800 N$

7 0

3 years ago

Read 2 more answers

What controls whether a lava flow is aa or pahoehoe

nevsk [136]

Explanation:

Basaltic lava

Basaltic lava generally takes two distinct forms known by the Hawaiian terms pahoehoe and aa. Pahoehoe has a smooth wavy surface that resembles twisted rope. It advances by extruding molten toes of lava beneath a thin, flexible crust. As it travels pahoehoe lava often changes to blocky flows called aa.

7 0

3 years ago

Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d

Tasya [4]

Answer:

I(x) = 1444×k × ${\pi}$

I(y) = 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$

Explanation:

Given data

function = x^2 + y^2 ≤ 36

function = x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to $\pi /2$ for θ

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ y²p dA

I(x) = $\int_{0}^{6}\int_{0}^{\pi/2}$ (a sinθ)²(k × a²) adθda

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (sin²θ)dθ

I(x) = k $\int_{0}^{6}a^(5)$ da × $\int_{0}^{\pi/2}$ (1-cos2θ)/2 dθ

I(x) = k $({r}^{6}/6)^(5)_0$ × ${θ/2 - sin2θ/4}^{\pi /2}_0$

I(x) = k × $({6}^{6}/6)$ × ( ${\pi /4} - sin\pi /4)$

I(x) = k × $({6}^{5})$ × ${\pi /4}$

I(x) = 1444×k × ${\pi}$ .....................1

and we can say I(x) = I(y) by the symmetry rule

and here I(o) will be I(x) + I(y) i.e

I(o) = 2 × 1444×k × ${\pi}$

I(o) = 3888×k × ${\pi}$ ......................2

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3 years ago

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