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2 years ago

100 points

Physics

1 answer:

luda_lava [24]2 years ago

5 0

Answer:

10.21m/s²

Explanation:

Radial acceleration, also known as centripetal acceleration, refers to the acceleration of an object along the circular radius. It can be calculated using the formula:

a(r) = v²/r

Where a(r) = radial acceleration

v = velocity

r = radius of the circular path

According to the information provided in the question, velocity (v) = 17.5m/s, radius = 30m. Hence;

a(r) = v²/r

a(r) = 17.5²/30

a(r) = 10.208333

a(r) = 10.21m/s²

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What is the net work doneon the object over the distance shown?

GuDViN [60]

$A)F_0d$

Explanation

If you graph the force on an object as a function of the position of that object, then the area under the curve will equal the work done on that object, so we need to find the area under the function to find the work

Step 1

find the area under the function.

Area:

$\text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red}$ $\begin{gathered} \text{the area of a rectangle is given by} \\ A_{rec}=lenght\cdot widht \\ \text{and} \\ \text{the area of a triangle is given by:} \\ A_{tr}=\frac{base\cdot height}{2} \end{gathered}$

$\begin{gathered} \text{Area}=rec\tan gle_{green}+triangle_{gren}-triangle_{red} \\ \text{replace} \\ \text{Area}=(F_0\cdot d)+\frac{(F_0\cdot d)}{2}-\frac{(F_0\cdot d)}{2} \\ \text{Area}=(F_0\cdot d) \\ Area=F_0d \end{gathered}$

therefore, the answer is

$A)F_0d$

I hope this helps you

4 0

9 months ago

If 30.45 grams of water is to be heated up 3.3 degrees to make baby

KengaRu [80]

Answer:

A. 420 J

Explanation:

Given the following data;

Mass = 30.45 g

Specific heat capacity = 4.18 J/g °C.

Temperature = 3.3°C

To find the quantity of heat;

Heat capacity is given by the formula;

$Q = mct$

Where;

Q represents the heat capacity or quantity of heat.

m represents the mass of an object.

c represents the specific heat capacity of water.

t represents the temperature.

Substituting into the equation, we have;

$Q = 30.45 * 4.18 * 3.3$

Q = 420.03 ≈ 420 Joules

8 0

2 years ago

A revolutionary war cannon, with a mass of 2260 kg, fires a 15.5 kg ball horizontally. The cannonball has a speed of 109 m/s aft

Anton [14]

Answer:

The gain in velocity is 0.37m/s

Explanation:

We need solve this problem though the conservation of momentum. That is,

$m_1 v_1 = m_2 v_2$

$m_1=2260Kg\\m_2=15.5Kg\\v_2= 109m/s$

Using the equation to find $v_1$ ,

$v_1=\frac{m_2 v_2}{m_1}\\v_1=\frac{15.5*109}{2260}\\v_1= 0.7475$

Using the conservation of energy equation, we have,

$KE= \frac{1}{2}m*v^2$

$KE_{cball}=\frac{1}{2}(15.5)(109)^2=92077.75J$

$KE_{cannon}=\frac{1}{2}(2260)(0.7475)^2=631.39J$

$Total KE= 92077.75+13425530=92708.9J$

Now this energy over the cannonball

$KE=\frac{1}{2}m*v_2^2$

$92708.9=\frac{1}{2}15.5v_2^2$

$V_2 = 109.37m/s$

The gain in velocity is 0.37m/s

4 0

3 years ago

Determine a formula for the magnitude of the force F exerted on the large block (Mc) so that the mass Ma does not move relative

SVEN [57.7K]

Answer:

The magnitude of the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *( $M_{b}$ *g/( $\sqrt{M_{a} ^{2}-M_{b} ^{2}}$ ))

Explanation:

Given there are three blocks of masses $M_{a}$ , $M_{b}$ and $M_{c}$ (ref image in attachment)

When all three masses move together at an acceleration a, the force F is given by

F = ( $M_{a}$ + $M_{b}$ + $M_{c}$ ) *a ................(equation 1)

Also it is given that $M_{a}$ does not move with respect to $M_{c}$ , which gives tension T is exerted on pulley by $M_{a}$ only, Hence tension T is

T = $M_{a}$ *a ..........(equation 2)

There is also also tension exerted by $M_{b}$ . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by

T = $M_{b}$ $\sqrt{a^{2} +g^{2} }$ ................(equation 3)