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3 years ago

13

Calculate the force exerted on the wall assuming that force is horizontal and using the data in the schematic representation of

the situation. Note that the force exerted on the wall is equal in magnitude and opposite in direction to the force exerted on the horse, keeping it in equilibrium. The total mass of the horse and rider is 675 kg. Take the data to be accurate to three digits.

1 answer:

Jlenok [28]3 years ago

8 0

Answer:

1.93 x 10∧3 N

Explanation:

The picture attached shows the calculation

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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

so

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

3 0

3 years ago

Assume that two of the electrons at the negative terminal have attached themselves to a nearby neutral atom. There is now a nega

devlian [24]

Answer: its negative

Explanation: becase it is

5 0

3 years ago

Water has a high heat capacity. What does this term mean? a) Water is relatively resistant to an increase in temperature. b) Wat

Delicious77 [7]

Answer:a) Water is relatively resistant to an increase in temperature

Explanation:

Water has relatively high heat capacity i.e. it requires high amount of heat to raise a little increase in temperature.

Heat capacity is defined as ability of molecule to absorb heat . Mathematically heat capacity is defined as

$C=\frac{Q}{\Delta T}$

where $C=heat\ capacity$

$Q=heat\ interaction$

$\Delta T=change\ in\ Temperature$

4 0

3 years ago

Chegg ) Alice owns 20 grams of a radioactive isotope that has a half-life of ln(4) years. (a) Find an equation for the mass m(t)

stepladder [879]

Answer:

$m(t)=20e^{-0.5t}$

Explanation:

Given:

Initial mass of isotope (m₀) = 20 g

Half life of the isotope $(t_{1/2})$ = (ln 4) years

The general form for the radioactive decay of a radioactive isotope is given as:

$m(t)=m_0e^{-kt}$

Where,

$m(t)\to mass\ after\ 't'\ years\\t\to years\ passed\\k\to rate\ of\ decay\ per\ year$

So, the equation is: $m(t)=20e^{-kt}$

At half-life, the mass is reduced to half of the initial value.

So, at $t=t_{1/2},m(t)=\frac{m_0}{2}$ . Plug in these values and solve for 'k'. This gives,

$\frac{m_0}{2}=m_0e^{-k\times\ln 4}\\\\0.5=e^{-k\times\ln 4}\\\\Taking\ natural\ log\ on\ both\ sides,we\ get:\\\\\ln(0.5)=-k\times \ln 4\\\\k=\frac{\ln 0.5}{-\ln 4}=0.5$

Hence, the equation for the mass remaining is given as:

$m(t)=20e^{-0.5t}$

8 0

3 years ago

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the pla

Mrrafil [7]

Answer:

F'=708.53 N

Explanation:

We have,

The lifting force, F, exerted on an airplane wing varies jointly as the area, A, of the wing's surface and the square of the plane's velocity, v. It means tat,

$F=kAv^2$

k is constant

If, A = 190 Ft², v = 220 mph, F = 950 pounds

Let's find k first from above data. So,

$k=\dfrac{F}{Av^2}\\\\k=\dfrac{950}{190\times 220^2}\\\\k=0.0001033$

It is required to find the lifting force on the wing if the plane slows down to 190 miles per hour. Let F' is the new force. So,

$F'=0.0001033\times 190\times (190)^2\\\\F'=708.53\ \text{pounds}$

So, the lifting force is 708.53 pounds if the plane slows down to 190 miles per hour.

3 0

3 years ago