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3 years ago

Help please! Will put brainliest

Mathematics

1 answer:

nata0808 [166]3 years ago

4 0

Answer:

Step-by-step explanation:

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The answer please please the answer

notsponge [240]

Answer:

25.5 sq cm

Step-by-step explanation:

1*3/2+7*3+1/2*3*2=3/2+21+3=25.5 sq cm

6 0

3 years ago

if 11,100 is invested at 3.1 percent interest compounded monthly, how much will the investment be worth in 19 years?

Fynjy0 [20]

The formula for compound interest is:
A=P(1+r/n)^(nt)
Where A represents the amount of money in the account after t years, P is the principal (investment), n is the number of compoundings per year, and r is the interest rate in decimal form.
P=11,100
r=.031
n=12 (monthly)
t=19
A=11,100(1+.031/12)^(12*19)
A=11,100(1+. 002583)^(228)
A=11,100(1.002583)^(228)
A=11,100(1.80082)
A=$19,989.10

3 0

4 years ago

Read 2 more answers

A worker drags a box of mass 50 kg across a horizontal floor. The worker attaches a rope to the box and pulls on the rope so tha

Arlecino [84]

Answer:

A) $F=\displaystyle{\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)} }$ with $a = \frac{2\Delta x}{t^2}$

B) The magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box

Step-by-step explanation:

A) The forces acting on the box are the pulling force that the worker exerts on the rope, the friction, the normal force, and the gravitational force. See the attached diagram. Since the pulling force on the rope makes an angle with the horizontal, this will have components in the x and y-axis (see diagram).

In the y-axis, the box does not move, therefore:

$\sum_{y} F = 0\\F\sin(\alpha) + N - F_g = 0\\N = F_g - F\sin(\alpha)\\N = mg - F\sin(\alpha)$ (1)

where $\alpha$ is the given angle of the rope with the horizontal.

In the x-axis, the box moves with an acceleration $a$ that can be calculated as a function of the given displacement and time interval as:

$a = \frac{2\Delta x}{t^2}$ and the force equation is:

$\sum_{x} F = ma\\F\cos(\alpha) -f =ma\\F\cos(\alpha) - \mu N = ma$

now substituting $N$ from equation (1):

$F\cos(\alpha) - \mu N = ma\\F\cos(\alpha) - \mu (mg - F\sin(\alpha))=ma\\F\cos(\alpha) - \mu mg -\mu F\sin(\alpha)=ma\\F\cos(\alpha) -\mu F\sin(\alpha)=ma+ \mu mg\\F(\cos(\alpha) -\mu \sin(\alpha))=ma+ \mu mg\\\boxed{F=\frac{m(a+ \mu g)}{\cos(\alpha) -\mu \sin(\alpha)}}$

and putting the expression for the acceleration:

$\boxed{F=\frac{m \frac{2\Delta x}{t^2}+ \mu mg}{\cos(\alpha) -\mu \sin(\alpha)}}$ (2)

which is the requested expression

B) Substituting the values in (2) and using $g=9.8\ \rm{ms^{-2}}$ :

$F=\frac{50 \cdot ( 0.762939+ 0.1\cdot \cdot 9.8)}{\cos(36.8^{\circ}) -0.1 \sin(36.8^{\circ})}\\F=\frac{87.147}{0.740829} \\F=11.634\ \rm{N}$

The gravitational force acting on the box is:

$F_g=mg=50\cdot9.8=490\ \rm{N}$

$F/F_g = (117.634/490)\cdot 100 = 24 \%$

Thus, the magnitude of the pulling force is 24 % of the magnitude of the gravitational force acting on the box.

8 0

4 years ago

Q7 Q16.) Convert the polar equation to a rectangular equation.

Mekhanik [1.2K]

In polar coordinates, we have

$x=r\cos\theta\implies\cos\theta=\dfrac xr$
$y=r\sin\theta\implies\sin\theta=\dfrac yr$

From this we can write

$\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{\frac yr}{\frac xr}=\dfrac yx$

If $\theta=\dfrac{4\pi}3$ , then $\tan\theta=\sqrt$ , so we get

$\dfrac yx=\sqrt3\implies y=\sqrt3\,x$

4 0

3 years ago

A weight suspended by a spring vibrates vertically according to the function D D given by D(t)=2sin(4π(t+18)) D ( t ) = 2 sin (

STALIN [3.7K]

Answer:

21.75 cm/s

Step-by-step explanation:

1 cm above central position means D = 1, so we plug in 1 into D(t) and find the time and which this occurs.

$D(t)=2Sin (4\pi(t+18))\\1=2Sin (4\pi(t+18))\\\frac{1}{2}=Sin(4\pi t +72\pi)\\4\pi t + 72 \pi = \frac{\pi}{6}\\4\pi t = \frac{\pi}{6}-72\pi\\t=\frac{\frac{\pi}{6}-72\pi}{4\pi}$

This is the time at which this occurs.

To find instantaneous rate of change, we differentiate D(t) and plug in this t we found. Remembering that d/dt (Sin t) = Cos t

$D(t)=2Sin (4\pi(t+18))\\D(t)=2Sin(4\pi t + 72\pi)\\D'(t)=2Cos(4\pi t + 72\pi)(4\pi)$

Now putting t:

$D'(t)=2Cos(4\pi t + 72\pi)(4\pi)\\D'(\frac{\frac{\pi}{6}-72\pi}{4\pi})=2Cos(4\pi (\frac{\frac{\pi}{6}-72\pi}{4\pi}) + 72\pi)(4\pi)\\=8\pi Cos(\frac{\pi}{6})\\=21.75$

Thus, the instantaneous rate would be around 21.75 cm/s

8 0

3 years ago