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2 years ago

10

Which type of radio active decays

1 answer:

stepan [7]2 years ago

3 0

Negative beta decay. So probably 4.

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Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed o

natulia [17]

Answer

given,

Speed of car A = 95 Km/h

= 95 x 0.278 = 26.41 m/s

Speed of Car B = 121 Km/h

= 121 x 0.278 = 33.64 m/s

Distance between Car A and B at t=0 = 41 Km

a) Distance travel by car B

d = 26.41 t + 41000

speed of the car A = 33.64 m/s

distance = s x t

26.41 t + 41000 = 33.64 x t

7.23 t = 41000

t = 5670.82 s

time taken by Car B to cross Car A is equal to t = 5670.82 s

distance traveled by car A

D = s x t = 26.41 x 5670.82 = 149766.25 m = 149.76 Km

b) distance travel by the car B in 30 s after overtaking car A

D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km

8 0

3 years ago

Which of the following is not a part of dalton s atomic theory?

Alex777 [14]

<span>c. atoms are always in motion..............</span>

6 0

3 years ago

Read 2 more answers

A wave has a wavelength of 20 mm and a frequency of 5 Hz what is the speed?

yKpoI14uk [10]

The answer is 100mm/s. I hope this helps :)

7 0

3 years ago

To develop muscle tone, a woman lifts a 2.50 kg weight held in her hand. She uses her biceps muscle to flex the lower arm throug

Romashka [77]

To solve this problem we will use the concepts related to Torque as a function of the Force in proportion to the radius to which it is applied. In turn, we will use the concepts of energy expressed as Work, and which is described as the Torque's rate of change in proportion to angular displacement:

$\tau = Fr$

Where,

F = Force

r = Radius

Replacing we have that,

$\tau = Fr$

$\tau = 21cm (\frac{1m}{100cm})* 550N$

$\tau = 11.55Nm$

The moment of inertia is given by 2.5kg of the weight in hand by the distance squared to the joint of the body of 24 cm, therefore

$I = 0.25Kg\cdot m^2 +(2.5kg)(0.24m)^2$

$I = 0.394kg\cdot m^2$

Finally, angular acceleration is a result of the expression of torque by inertia, therefore

$\tau = I\alpha \rightarrow \alpha = \frac{\tau}{I}$

$\alpha = \frac{11.55}{0.394}$

$\alpha = 29.3 rad/s^2$

PART B)

The work done is equivalent to the torque applied by the distance traveled by 60 °° in radians $(\pi / 3)$ , therefore

$W = \tau \theta$

$W = 11.5* \frac{\pi}{3}$

$W = 12.09J$

4 0

3 years ago

To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d

Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is $v = 350 \ m/s$

Explanation:

From the question we are told that

The distance of separation is d = 4.00 m

The distance of the listener to the center between the speakers is I = 5.00 m

The change in the distance of the speaker is by $k = 60 cm = 0.6 \ m$

The frequency of both speakers is $f = 700 \ Hz$

Generally the distance of the listener to the first speaker is mathematically represented as

$L_1 = \sqrt{l^2 + [\frac{d}{2} ]^2}$

$L_1 = \sqrt{5^2 + [\frac{4}{2} ]^2}$

$L_1 = 5.39 \ m$

Generally the distance of the listener to second speaker at its new position is

$L_2 = \sqrt{l^2 + [\frac{d}{2} ]^2 + k}$

$L_2 = \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}$

$L_2 = 5.64 \ m$

Generally the path difference between the speakers is mathematically represented as

$pD = L_2 - L_1 = \frac{n * \lambda}{2}$

Here $\lambda$ is the wavelength which is mathematically represented as

$\lambda = \frac{v}{f}$

=> $L_2 - L_1 = \frac{n * \frac{v}{f}}{2}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

=> $L_2 - L_1 = \frac{n * v}{2f}$

Here n is the order of the maxima with value of n = 1 this because we are considering two adjacent waves

=> $5.64 - 5.39 = \frac{1 * v}{2*700}$

=> $v = 350 \ m/s$

7 0

4 years ago