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daser333 [38]
3 years ago
7

PLEASEEE HElPP my time is almost up and I need help!!!!

Physics
1 answer:
Ray Of Light [21]3 years ago
5 0
WARNING!!! IF YOU SEE BOTS ASKING YOU TO GO TO A SITE CALLED BRAINLY. TODAY, DONT GO! THEY ARE USING THE BRAINLY NAME TO MAKE A FAKE WEBSITE LINK TO A INAPPROPRIATE WEBSITE THAT HAS NOTHING TO DO WITH BRAINLY! IF YOU SEE ANY OF THESE BOTS AVOID THEM!!! THEY TEND TO HAVE LINKS OF PICTURES THAT ARENT ACTUALLY THEM! THEY ARE JUST PROMOTING A INAPPROPRIATE WEBSITE THAT ISNT MEANT FOR KIDS LIKE US!
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The body falls in a free fall 11s.Calculate: a) from what height the body of the paddle) what path did it fall during the last s
JulsSmile [24]

Answer:

a) h = 593.50 m

b) h₁₁ = 103 m

c) vf = 107.91 m/s

Explanation:

a)

We will use second equation of motion to find the height:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height = ?

vi = initial speed = 0 m/s

t = time taken = 11 s

g = 9.81 /s²

Therefore,

h = (0\ m/s)(11\ s) + (\frac{1}{2})(9.8\ m/s^2)(11\ s)^2\\\\

<u>h = 593.50 m</u>

b)

For the distance travelled in last second, we first need to find velocity at 10th second by using first equation of motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = v₁₀ = ?

t = 10 s

vi = 0 m/s

Therefore,

v_{10} = 0\ m/s + (9.81\ m/s^2)(10\ s)\\\\v_{10} = 98.1\ m/s

Now, we use the 2nd equation of motion between 10 and 11 seconds to find the height covered during last second:

h = v_{i}t + (\frac{1}{2})gt^2

where,

h = height covered during last second = h₁₁ =  ?

vi = v₁₀ = 98.1 m/s

t = 1 s

Therefore,

h_{11} = (98.1\ m/s)(1\ s) + (\frac{1}{2})(9.8\ m/s^2)(1\ s)^2\\\\

<u>h₁₁ = 103 m</u>

c)

Now, we use first equation of motion for complete motion:

v_{f} = v_{i} + gt

where,

vf = final velocity at tenth second = ?

t = 11 s

vi = 0 m/s

Therefore,

v_{f} = 0\ m/s + (9.81\ m/s^2)(11\ s)

<u>vf = 107.91 m/s</u>

8 0
4 years ago
Positive psychologists believe that happiness can be increased by
Elenna [48]
The answer is positive thoughts
5 0
3 years ago
Read 2 more answers
A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?
zloy xaker [14]

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

<u>I = 4 A</u>

8 0
3 years ago
Two blocks slide without friction. Block 1 has a mass of 1.6 kg and a velocity of +5.5 m/s. Block 2 has a mass of 2.4 kg and a v
fomenos

Answer:

The velocity of block 1 after the collision is 1.9 m/s.

Explanation:

Hi there!

The momentum of the system composed by the two blocks is conserved (i.e. it remains constant) because no external force is acting on the blocks at the moment of the collision. Then, the momentum of the system before the collision is equal to the momentum after the collision. The momentum of the system is calculated adding the momenta of the two blocks.

initial momentum of the system = final momentum of the system

p1 + p2 = p1´ + p2´

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

Where:

p1 = initial momentum of block 1.

p2 = initial momentum of block 2.

p1´ = final momentum of block 1.

p2´ = final momentum of block 2.

m1 = mass of block 1.

m2 = mass of block 2.

v1 = initial velocity of block 1 (before the collision).

v2 = initial velocity of block 2.

v1´ = final velocity of block 1.

v2´ = final velocity of block 2.

Let´s write the equation with the data we have:

m1 · v1 + m2 · v2 = m1 · v1´ + m2 · v2´

1.6 kg · 5.5 m/s + 2.4 kg · 2.5 m/s = 1.6 kg · v1´ + 2.4 kg · 4.9 m/s

Solving for v1´:

14.8 kg · m/s = 1.6 kg · v1´ + 11.76 kg · m/s

(14.8 kg · m/s - 11.76 kg · m/s) / 1.6 kg = v1´

v1´ = 1.9 m/s

The velocity of block 1 after the collision is 1.9 m/s.

6 0
4 years ago
A force in the + x-direction with magnitude F(x) = 18.0N -(0.530 N/m)x is applied to a 6.00-kg box that is sitting on the horizo
gogolik [260]

Answer:v=8.17 m/s

Explanation:

Given

F(x)=18-0.530 x

mass of box=6 kg

distance traveled is 14 m

Work done by force

W=\int_{0}^{14}Fdx

W=\int_{0}^{14}(18-0.53x)dx

W=\left ( 18x-\frac{0.530}{2}x^2\right )_0^{14}

W=252-51.94=200.06 J

Work done by the force is equal to change in kinetic energy

W=\frac{mv^2}{2}

200.06\frac{6\times v^2}{2}

v=8.17 m/s

5 0
4 years ago
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