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Andrew [12]
3 years ago
12

Name two different types of precipitation?

Physics
2 answers:
sertanlavr [38]3 years ago
5 0

ok so the two types of precipitation are rain and snow

Norma-Jean [14]3 years ago
4 0

Answer:

2 different types of precipitation is rain and snow.

Explanation:

The clouds will form... and the droplets that could be coming out is rain and snow.

You might be interested in
Tiles are 6 mm long how many can you fit into a space 54 cm long
vlabodo [156]
It fits 90 times because 54÷6= 9 so if 54÷0.6 it will equal 90

4 0
3 years ago
Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 5.30 m/
Tju [1.3M]

Answer:

a) v_{p}  = 2.83 m / s ,  b)  50.5º north east

Explanation:

This is a vector problem.

         v_{bg} = v_{bm} + v_{mg}

The speed of the ball with respect to the ground is the speed of the ball with respect to Mia plus the speed of Mia with respect to the ground

To make the sum we decompose the speed of the ball in its components

The angle of 30 east of the south, measured from the positive side of the x axis is

             θ = 30 + 270 = 300

            v_{bx} = v_{b} cos 300

             v_{by} = v_{b} sin. 300

             v_{bx} = 3.60 cos 300 = 1.8 m / s

             v_{by} = 3.60 sin 300 = -3,118 m / s

Let's add speeds on each axis

X axis

            vₓ = v_{bx}

             vₓ = 1.8 m / s

Y Axis  

             v_{y} = v1 - vpy

             v_{y} = 5.30 - 3.118

             v_{y} = 2.182 m / s

The magnitude of the velocity can be found using the Pythagorean theorem

              v_{p} = √ (vₓ² + v_{y}²)

               v_{p} = √ (1.8² + 2.182²)

               v_{p} = 2,829 m / s

               v_{p}  = 2.83 m / s

b) for direction use trigonometry

              tan θ = v_{y} / vₓ

              θ = tan ⁺¹ v_{y} / vₓ

              θ = tan⁻¹ 2.182 / 1.8

         Tea = 50.48º

This address is 50.5º north east

8 0
2 years ago
Sarah weighs 392 N. She pushes a 56 kg rock with a force of 65 N. What force does the rock exert on Sarah?
Leona [35]

Answer:

The force the rock exerts on Sarah =#is 65 N

Explanation:

The given parameters are;

Sarah's weight = 392 N

The force with which Sarah pushes the rock = 65 N

The mass of the rock = 56 kg

The weight of the rock = The mass of the rock × Acceleration due to gravity

∴ The weight of the rock = 56 kg ×9.81 m/s² = 549.36 N

Given that the force Sarah applies to push the rock = 65 N, then by Newton's third law of motion which states that action and reaction are equal and opposite, the force that the rock exert on Sarah is equal an opposite to the force Sarah is applying

Therefore, the force the rock exerts on Sarah = 65 N (in the opposite direction).

6 0
2 years ago
A 2.40 cm × 2.40 cm square loop of wire with resistance 1.20×10−2 Ω has one edge parallel to a long straight wire. The near edge
Norma-Jean [14]

Answer:

current in loops is 52.73 μA

Explanation:

given data

side of square a = b  = 2.40 cm = 0.024 m

resistance R = 1.20×10^−2 Ω

edge of the loop c  = 1.20 cm = 0.012 m

rate of current = 120 A/s

to find out

current in the loop

solution

we know current formula that is

current = voltage / resistance    .................a

so current = 1/R × d∅/dt

and we know here that

flux ∅ = ( μ×I×b / 2π ) × ln (a+c/c)    ...............b

so

d∅/dt = ( μ×b / 2π ) × ln (a+c/c) × dI/dt       ...........c

so from equation a we get here current

current = ( μ×b / 2πR ) × ln (a+c/c) × dI/dt

current = ( 4π×10^{-7}×0.024 / 2π(1.20×10^{-2}) × ln (0.024 + 0.012/0.012) × 120

solve it and we get current that is

current = 4 ×10^{-7}× 1.09861 × 120

current = 52.73 ×10^{-6}  A

so here current in loops is 52.73 μA

8 0
2 years ago
A block of aluminum at a temperature of T1 = 32 degrees C has a mass of m1 = 12 kg. It is brought into contact with another bloc
Mariulka [41]

Answer:T=21.33 ^{\circ}

Explanation:

Given

mass of First Block m_1=12 kg

Temperature T_1=32^{\circ}

mass of second block m_2=0.5 m_1=6 kg

Temperature T_2=9^{\circ}

Heat capacity of aluminium c=899 J/kg-K

Final Temperature acquired by both blocks at steady state

Heat loss first block =Heat gain by second block

m_1\times c\times (32-T)=m_2\times c\times (T-9)

12\times 899\times (32-T)=6\times 899\times (T-9)

2\times 32=3T

T=\frac{64}{3}=21.33 ^{\circ}

5 0
3 years ago
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