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Dennis_Churaev [7]
3 years ago
15

Pretty please help yah girl out :)

Physics
2 answers:
Rudik [331]3 years ago
5 0
The answer is c reflection hope this helps :)
umka2103 [35]3 years ago
5 0
I would have said A.
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An electric field around two charged objects is shown.
mamaluj [8]

Answer: D

<u></u>

X: positive

Y: negative

Explanation:

It's either A or D and I chose A and got it wrong.

3 0
3 years ago
Q3. A man walks with a speed of 8 m/sec in 40 sec. How much distance was covered?
chubhunter [2.5K]

Answer:

it is320 m.

Explanation:

by formula

d= m/s×t

d= 8× 40

d=320 m

8 0
2 years ago
Read 2 more answers
A 2 m tall, 0.5 m inside diameter tank is filled with water. A 10 cm hole is opened 0.75 m from the bottom of the tank. What is
valina [46]

Answer:

4.75 m/s

Explanation:

The computation of the velocity of the existing water is shown below:

Data provided in the question

Tall = 2 m

Inside diameter tank = 2m

Hole opened = 10 cm

Bottom of the tank = 0.75 m

Based on the above information, first we have to determine the height which is

= 2 - 0.75 - 0.10

= 2 - 0.85

= 1.15 m

We assume the following things

1. Compressible flow

2. Stream line followed

Now applied the Bernoulli equation to section 1 and 2

So we get

\frac{P_1}{p_g} + \frac{v_1^2}{2g} + z_1 = \frac{P_2}{p_g} + \frac{v_2^2}{2g} + z_2

where,

P_1 = P_2 = hydrostatic

z_1 = 0

z_2 = h

Now

\frac{v_1^2}{2g} + 0 = \frac{v_2^2}{2g} + h\\\\V_2 < < V_1 or V_2 = 0\\\\Therefore\  \frac{v_1^2}{2g} = h\\\\v_1^2 = 2gh\\\\ v_1 = \sqrt{2gh} \\\\v_1 = \sqrt{2\times 9.8\times 1.15}

= 4.7476 m/sec

= 4.75 m/s

6 0
3 years ago
Find an equation of the line which is parallel to 2x-3y=6 and passes through the point (6,9)
tatuchka [14]
Step 2: Use the slope to find<span> the y-intercept. </span>Line<span> is </span>parallel<span> so use m = 2/5. </span>6<span>. </span>Find<span>the </span>equation<span> of a </span>line passing through the point<span> (8, –</span>9<span>) perpendicular to the </span>line<span> 3x + 8y = 4.</span>
8 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

V_f= Final Velocity

Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
3 years ago
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