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Helen [10]
3 years ago
14

When a certain isotope, such as U-238, is hit by a neutron, it will always split into the same smaller nuclei.

Physics
2 answers:
nadezda [96]3 years ago
4 0

Answer: falss

Explanation:

il63 [147K]3 years ago
4 0

Answer:

false

Explanation:

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Which feature of electromagnets makes them more useful than permanent
mezya [45]
I think it’s D................
4 0
3 years ago
A light spring having a force constant of 115 N/m is used to pull a 9.00 kg sled on a horizontal frictionless ice rink. The sled
Ksivusya [100]

Answer:

Stretch in the spring = 0.1643 (Approx)

Explanation:

Given:

Mass of the sled (m) = 9 kg

Acceleration of the sled (a) = 2.10 m/s ²

Spring constant (k) = 115 N/m

Computation:

Tension force in the spring  (T) = ma

Tension force in the spring  (T) = 9 × 2.10

Tension force in the spring  (T) = 18.9 N

Tension force in the spring = Spring constant (k) × Stretch in the spring

18.9 N = 115 N  × Stretch in the spring

Stretch in the spring = 18.9 / 115

Stretch in the spring = 0.1643 (Approx)

8 0
3 years ago
Infrared light of wavelength 2.5 µm illuminates a 0.20-mm-diameter hole. What is the angle of the first dark fringe in radians?
stepan [7]

Answer:

Θ=0.01525 rad

or

Θ=0.87°

Explanation:

Given data

wavelength λ=2.5 µm =2.5×10⁻⁶m

Diameter d=0.20 mm =0.20×10⁻³m

To find

Angle Θ in radians and degree

Solution

Circular apertures have first dark fringe at

Θ=(1.22λ)/d

Substitute the given values

So

Θ=[1.22(2.5×10⁻⁶m)]/0.20×10⁻³m

Θ=0.01525 rad

or

Θ=0.87°

6 0
3 years ago
Goat gallops at -1.5m/s, how far will the goat travel in 20 seconds?<br><br> Please show work
saw5 [17]

Answer:

30m

Explanation:

Given parameters:

Speed of the goat  = 1.5m/s

Time  = 20s

Unknown:

Distance traveled  = ?

Solution:

To solve this problem, we use;

      Speed  = \frac{distance}{time}  

 Distance = Speed x time

 So;

 Distance  = 1.5m/s x 20s  = 30m

5 0
3 years ago
The Moon (7.35 x 10^22 kg) is 3.85 x 10^8 m from the Earth and orbits around it in a circle. a) If the period of the Moon is 27.
Lemur [1.5K]

a) Speed of the Moon: 1025 m/s

The speed of the moon is equal to the ratio between the circumference of its orbit and the orbital period:

v=\frac{2 \pi r}{T}

where

r=3.85\cdot 10^8 m is the radius of the orbit of the Moon

T=27.3 d \cdot (24 h/d) \cdot (60 min/h) \cdot (60 s/min)=2.36\cdot 10^6 s is the orbital period

Substituting into the formula, we find

v=\frac{2 \pi (3.85\cdot 10^8 m)}{2.36\cdot 10^6 s)}=1025 m/s


b) Centripetal force: 2.0 \cdot 10^{20} N

The centripetal force acting on the Moon is given by:

F=m\frac{v^2}{r}

where

m=7.35 \cdot 10^{22}kg is the mass of the Moon

v=1025 m/s is its orbital speed

r=3.85\cdot 10^8 m is the radius of the orbit

Substituting into the formula, we find

F=(7.35 \cdot 10^{22} kg) \frac{(1025 m/s)^2}{3.85\cdot 10^8 m}=2.0 \cdot 10^{20} N


c) Gravitational force

The only relevant force that acts on the Moon, and that keeps the Moon in circular motion around the Earth, is the gravitational force exerted by the Earth on the Moon. In fact, this force "pulls" the Moon towards the Earth, so towards the centre of the orbit of the Moon, therefore it acts as source of centripetal force for the Moon.

8 0
3 years ago
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