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3 years ago

An ice cream truck is going 25m/s to East. It accelerates to 40m/s in the same direction over 3s. What is its acceleration

2 answers:

lina2011 [118]3 years ago

8 0

40 m/s subtracted by 25m/s is 15.
15 has to be divided by the additional 3s which is 5
The 5 miles and the seconds left behind are 2
Answer: 5m/s2

skad [1K]3 years ago

4 0

Its acceleration is 5m/s2.
The formula for acceleration is Final speed - initial speed/time.
So 40m/s - 25m/s is 15m/s.
15m/s divided by 3s is 5m/s2.
Thus the acceleration is 5m/s2

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A uniform solid sphere has mass m= 7 kg and radius r= 0. 4 m. What is its moment of inertia about an axis tangent to its surface

lilavasa [31]

The moment of inertia of a uniform solid sphere is equal to 0.448 $kgm^2$ .

Given the following data:

Mass of sphere = 7 kg.

Radius of sphere = 0.4 meter.

<h3>How to calculate moment of inertia.</h3>

Mathematically, the moment of inertia of a solid sphere is given by this formula:

$I=\frac{2}{5} mr^2$

Where:

I is the moment of inertia.

m is the mass.

r is the radius.

Substituting the given parameters into the formula, we have;

$I=\frac{2}{5} \times 7 \times 0.4^2\\\\I=2.8 \times 0.16$

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2 years ago

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Aleonysh [2.5K]

Answer:

What would most likely happen as a result of the generator in a wind turbine breaking?

The What would most likely happen as a result of the generator in a wind turbine breaking?

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3 years ago

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Answer:

<h2>Virtual image</h2>

Explanation:

<h3>Virtual image can be caught on a screen</h3>

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3 years ago

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A 1.60 m cylindrical rod of diameter 0.550 cm is connected to a power supply that maintains a constant potential difference of 1

bija089 [108]

Answer:

Part a)

$\rho = 1.35 \times 10^{-5}$

Part b)

$\alpha = 1.12 \times 10^{-3}$

Explanation:

Part a)

Length of the rod is 1.60 m

diameter = 0.550 cm

now if the current in the ammeter is given as

$i = 18.7 A$

V = 17.0 volts

now we will have

$V = I R$

$17.0 = 18.7 R$

R = 0.91 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.91 = \rho \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho = 1.35 \times 10^{-5}$

Part b)

Now at higher temperature we have

$V = I R$

$17.0 = 17.3 R$

R = 0.98 ohm

now we know that

$R = \rho \frac{L}{A}$

$0.98 = \rho' \frac{1.60}{\pi(0.275\times 10^{-2})^2}$

$\rho' = 1.46 \times 10^{-5}$

so we will have

$\rho' = \rho(1 + \alpha \Delta T)$

$1.46 \times 10^{-5} = 1.35 \times 10^{-5}(1 + \alpha (92 - 20))$

$\alpha = 1.12 \times 10^{-3}$

Answer:

Part a)

$i = 1.55 A$

Part b)

$v_d = 1.4 \times 10^{-4} m/s$

Explanation:

Part a)

As we know that current density is defined as

$j = \frac{i}{A}$

now we have

$i = jA$

Now we have

$j = 1.90 \times 10^6 A/m^2$

$A = \pi(\frac{1.02 \times 10^{-3}}{2})^2$

so we will have

$i = 1.55 A$

Part b)

now we have

$j = nev_d$

so we have

$n = 8.5 \times 10^{28}$

$e = 1.6 \times 10^{-19} C$

so we have

$1.90 \times 10^6 = (8.5 \times 10^{28})(1.6 \times 10^{-19})v_d$

$v_d = 1.4 \times 10^{-4} m/s$

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