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4 years ago

7

A -kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the tra

ck is meters. The magnitude of the car's centripetal acceleration is:_______

1 answer:

Tatiana [17]4 years ago

4 0

Answer:

The centripetal acceleration of the car is $8\ m/s^2$ .

Explanation:

Let the mass of the car, $m=10^3\ kg$

Diameter of the circular path, d = 100 m

Speed of car, v = 20 m/s

Radius, r = 50 m

When an object moves in a circular path, the centripetal acceleration acts on it. It is given by :

$a=\dfrac{v^2}{r}$

$a=\dfrac{(20\ m/s)^2}{50\ m}$

$a=8\ m/s^2$

So, the centripetal acceleration of the car is $8\ m/s^2$ . Hence, this is the required solution.

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A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft)

Phantasy [73]

Complete Question

A 100-W (watt) light bulb has resistance R=143Ω (ohms) when attached to household current, where voltage varies as V=V0sin(2πft), where V0=110 V, f=60 Hz. The power supplied to the bulb is P=V2R J/s (joules per second) and the total energy expended over a time period [0,T] (in seconds) is $U = \int\limits^T_0 {P(t)} \, dt$

Compute U if the bulb remains on for 5h

Answer:

The value is $U = 7.563 *10^{5} \ J$

Explanation:

From the question we are told that

The power rating of the bulb is $P = 100 \ W$

The resistance is $R = 143 \ \Omega$

The voltage is $V = V_o sin [2 \pi ft]$

The energy expanded is $U = \int\limits^T_0 {P(t)} \, dt$

The voltage $V_o = 110 \ V$

The frequency is $f = 60 \ Hz$

The time considered is $t = 5 \ h = 18000 \ s$

Generally power is mathematically represented as

$P = \frac{V^2}{ R}$

=> $P = \frac{( 110 sin [2 \pi * 60t])^2}{ 144}$

=> $P = \frac{ 110^2 [ sin [120 \pi t])^2}{ 144}$

So

$U = \int\limits^T_0 { \frac{ 110^2* [sin [120 \pi t])^2}{ 144}} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { ( sin^2 [120 \pi t]} \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 2 (120\pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} \int\limits^T_0 { \frac{1 - cos 240 \pi t)}{2} } \, dt$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | T} \atop {0}} \right.$

=> $U = \frac{110^2}{144} [\frac{t}{2} - [\frac{1}{2} * \frac{sin(240 \pi t)}{240 \pi} ] ]\left | 18000} \atop {0}} \right.$

$U = \frac{110^2}{144} [\frac{18000}{2} - [\frac{1}{2} * \frac{sin(240 \pi (18000))}{240 \pi} ] ]$

=> $U = 7.563 *10^{5} \ J$

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3 years ago

a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi

Mars2501 [29]

The radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

r/10 = tan [arctan(1/2)] = tan (1/2)

Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = $\sqrt{5} -2$

Hence, r = $10\sqrt{5} -20$

So, the radius of the sphere in meters is ,r = $10\sqrt{5} -20$

Learn more about radius (r) of the sphere here;

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1 year ago

20 N effort is used in the 20 cm long spanner to unscrew a nut, then calculate momentum to unscrew the nut. [ Ans=4 Nm]

irga5000 [103]

Answer:

Refer to the attachment.

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3 years ago

Convert 172 to a scientific notation

gtnhenbr [62]

Answer:

1.72×10^2

Explanation:

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3 years ago

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Potential difference of a battery is 2.2 V when it is connected

Alchen [17]

Answer:

1.1ohms

Explanation:

According to ohms law E = IR

If potential difference of a battery is 2.2 V when it is connected across a resistance of 5 ohm and if suddenly the voltage Falls to 1.8V then the current in the 5ohms resistor I = V/R = 1.8/5

I = 0.36A (This will be the load current).

Before we can calculate the value of the internal resistance, we need to know the voltage drop across the internal resistance.

Voltage drop = 2.2V - 1.8V = 0.4V

Then we calculate the internal resistance using ohms law.

According to the law, V = Ir

V= voltage drop

I is the load current

r = internal resistance

0.4 = 0.36r

r = 0.4/0.36

r = 1.1 ohms

6 0

4 years ago