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mestny [16]
3 years ago
15

How would the composition of an atom change if both the atomic number and mass number each increase by one? A) The atom would ha

ve 1 more neutron. B) The atom would have one more proton and electron. C) The atom would have one more proton and one more neutron. D) The atom would have one more proton, one more neutron and one more electron.
Physics
1 answer:
weqwewe [10]3 years ago
8 0
Choice-B is the correct one. 

-- The atomic number is the number of protons in the nucleus.
-- Each proton in the nucleus is usually matched by one electron in the 'cloud'.
-- The addition of a proton OR a neutron increases the mass number by 1 .
-- Electrons have such small mass that they don't figure into the atomic mass at all.
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If a 60-g object has a volume of 30 cm", what is its density?
alisha [4.7K]

Answer:

Density = Mass/volume. D= 60/30.Divide it and you'll get ur answer as 2

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3 years ago
Subduction occurs at which of the following tectonic plate boundaries?
exis [7]
When two tectonic plates collide and form a converging plate boundry, normally one of the plates will slide underneath the other and that is when Subduction occurs.
5 0
3 years ago
A hunter aims at a deer which is 40 yards away. Her cross- bow is at a height of 5ft, and she aims for a spot on the deer 4ft ab
shutvik [7]

Answer:

a)  θ₁ = 0.487º , b)   t = 0.400 s ,        x = 11.73 ft

Explanation:

For this exercise let's use the projectile launch relationships.

The initial height is I = 5 ft and the final height y = 4 ft

            y = y₀ + v_{oy} t - ½ g t²

The distance to the band is x = 40 yard (3 ft / 1 yard) = 120 ft

            x = v₀ₓ t

            t = x / v₀ₓ

We replace

             y –y₀ = v_{oy} x / v₀ₓ - ½ g x² / v₀ₓ²

             v_{oy} = v₀ sin θ

             v₀ₓ = vo cos θ

             

             y –y₀ = x tan θ - ½ g x² / v₀² cos² θ

                5-4 = 120 tan θ - ½ 32 120 / (300 2 cos2 θ)

                1 = 120 tan θ - 0.0213 sec² θ

Let's use the trigonometry relationship

               Sec² θ = 1 - tan² θ

                 1 = 120 tan θ - 0.0213 (1 –tan²θ)

                 0.0213 tan²θ + 120 tanθ -1.0213 = 0

                 

We change variables

          u = tan θ

          u² + 5633.8 u - 48.03 = 0

We solve the second degree equation

          u = [-5633.8 ±√(5633.8 2 + 4 48.03)] / 2

          u = [- 5633.8 ± 5633.82] / 2

           u₁ = 0.0085

           u₂= -5633.81

           u = tan θ

           θ = tan⁻¹ u

For u₁

           θ₁ = tan⁻¹ 0.0085

           θ₁ = 0.487º

For u₂

           θ₂ = -89.99º

The launch angle must be 0.487º

b) let's look for the time it takes for the arrow to arrive

         x = v₀ₓ t

         t = x / v₀ cos θ

         

         t = 120 / (300 cos 0.487)

         t = 0.400 s

The deer must be at a distance of

           v = 20 mph (5280 ft / 1 mi) (1 h / 3600s) = 29.33 ft / s

           x = v t

           x = 29.33 0.4

           x = 11.73 ft

3 0
3 years ago
What waves are used in the hospitals to take pictures of bones
stich3 [128]

Answer:

An X-ray is used to take pictures of your bones. The waves that are used are known as radiation waves.

7 0
3 years ago
A 7 kg ball is moving at a constant speed of 5 m/s. A force of 300 N is applied to the ball for 4 s. The new speed of the ball i
olasank [31]

Answer:

The new speed of the ball is 176.43 m/s

Explanation:

Given;

mass of the ball, m = 7 kg

initial speed of the ball, u = 5 m/s

applied force, F = 300 N

time of force action on the ball, t = 4 s

Apply Newton's second law of motion;

F = ma = \frac{m(v-u)}{t}\\\\m(v-u) = Ft\\\\v-u = \frac{Ft}{m}\\\\v =  \frac{Ft}{m} + u

where;

v is new speed of the ball

v =  \frac{Ft}{m} + u\\\\v =\frac{300*4}{7} + 5\\\\v = 176.43 \ m/s

Therefore, the new speed of the ball is 176.43 m/s

8 0
3 years ago
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