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chubhunter [2.5K]
3 years ago
8

What nuclear process changes one element into another different element?

Physics
2 answers:
klio [65]3 years ago
6 0
The answer would option C
Semmy [17]3 years ago
4 0
The answer is nuclear transmutation
You might be interested in
We know the frequency range of certain sounds are: 400-560 Hz, what are the ranges of wavelength in meters when the signal trans
Ksivusya [100]

Answer:

Range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

Explanation:

We have given range of frequency is 400-560 Hz

Speed of the light c=3\times 10^8m/sec

We have to find the range of the wavelength of signal transmitted

Ween know that velocity is given by v=\lambda f, here \lambda is wavelength and f is frequency

So for 400 Hz frequency wavelength will be \lambda =\frac{c}{f}=\frac{3\times 10^8}{400}=7.5\times 10^5m

And wavelength for frequency 560 Hz \lambda =\frac{c}{f}=\frac{3\times 10^8}{560}=5.35\times 10^5m

So range of wavelength will be 5.35\times 10^5m to 7.5\times 10^5m

8 0
3 years ago
Many people drink 2 cans of coke each day. If each can has 39 grams of sugar, how many pounds of sugar is consumed each week? Gi
trasher [3.6K]

Answer:

1.20372

Explanation: start with 39 times 2 for how much grams each day and then multiply that by 7 then the convert grams into pounds

5 0
3 years ago
The ability to clearly see objects at a distance but not close up is properly called ________. The ability to clearly see object
melamori03 [73]

Answer:

Hyperopia

Explanation:

     In hyperopia ,people face difficulties  to see close up object , but can see object easily which are at a distance.

The main reason of hyperopia is our eyeball.When our eyeball become too  short , then light focus behind the retina. Sowe will face problem to see near object but we can see distance object easily. Hyperopia is the opposite of nearsightedness. Hyperopia can be corrected by using contact lenses.

5 0
3 years ago
Can someone please help me with these physics problems? I just don’t even know where to start.
KIM [24]

#1

for the block of mass 5 kg normal force is given as

F_n = mg

F_n = 5*9.8 = 49 N

friction force is given as

F_f = \mu F_n

F_f = 0.1*49 = 4.9 N

Net force is given as

F_{net} = ma

F_{net} = 5*2 = 10 N

now we know that

F_{net} = F_{app} - F_f

10 = F_{app} - 4.9

F_{app} = 14.9 N

#2

Normal force is given as

F_n = mg

F_n = 6*9.8

F_n = 58.8 N

now we know that

F_{net} = F_{app} - F_f

F_{net} = 0

as object moves with constant velocity

F_{app} = F_f = 15 N

now for coefficient of friction we can use

F_f = \mu F_n

15 = \mu * 58.8

\mu = 0.255

#3

net force upwards is given as

F = 1.2 * 10^{-4} N

mass is given as

m = 7 * 10^{-5} kg

now as per newton's law we can say

F = ma

1.2 * 10^{-4} = 7 * 10^{-5} * a

a = 1.71 m/s^2

#4

As we know that when block is sliding on rough surface

part a)

net force = applied force - frictional force

F_{net} = F_{app} - F_f

ma = F_{app} - F_f

5*6 = 40 - F_{f}

F_f = 40 - 30 = 10 N

part b)

for coefficient of friction we can use

F_f = \mu F_n

10 = \mu * F_n

here normal force is given as

F_n = mg = 5*9.8 = 49 N

now we have

\mu = \frac{10}{49} = 0.204

#5

if an object is initially at rest and moves 20 m in 5 s

so we can use kinematics to find out the acceleration

d = v_i*t + \frac{1}{2}at^2

20 = 0 + \frac{1}{2}a(5^2)

a = 1.6 m/s^2

now net force is given as

F_{net} = ma

F_{net} = 10*1.6 = 16 N

#6

an object travelling with speed 25 m/s comes to stop in 1.5 s

so here acceleration of object is given as

a = \frac{v_f - v_i}{t}

a = \frac{0 - 25}{1.5} = -16.67 m/s^2

now the force is gievn as

F = ma

F = 5*16.67 = 83.3 N

3 0
4 years ago
40 POINTS!!!
Feliz [49]

1.) B.

2.) A.

3.) A.

If these are wrong please delete.

Thank You!

<3

4 0
3 years ago
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