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Virty [35]
3 years ago
11

If object is above it lowest point what energy does it have

Physics
1 answer:
spin [16.1K]3 years ago
7 0
I don’t understand the question can u reword it?
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A 0.155 kg arrow is shot upward
Soloha48 [4]

Answer: 244.05 J

Explanation:

To find speed  at 30 m above the ground use equation:

V²=Vo²-2Gs

V0=31.4m/s

s=30m

G=9.81m/s²

-----------------------

V²=31.4²-2*9.81*30

V²=985.96+588.6

V²=1574.56

V=39.68m/s ---speed of arrow on 30 m obove the ground

Use equation for kinetic enrgy:

Ke=mV²/2

m=0.155kg

V=39.68m/s

-------------------------

Ke=0.155kg*(39.68m/s)²/2

Ke=0.155*1574.5/2

Ke=244.05J

6 0
3 years ago
A straight vertical wire carries a current of 1.0 A into the page in a region where the magnetic field strength is 0.60 T. What
Advocard [28]

Answer:

force on the wire of section  cm will be 6\times 10^{-3}N

Direction of force on the wire will be in south direction        

Explanation:

We have given current in the wire i = 1 A

Magnetic field strength B = 0.6 T

We have to find the force on 1 cm section of the wire so l = 1 cm = 0.01 m

Force on the wire containing current is equal to

F=iBL

F=1\times 0.6\times 0.01=6\times 10^{-3}N

So force on the wire of section  cm will be 6\times 10^{-3}N

Direction of force on the wire will be in south direction

5 0
3 years ago
Does the world actually end on April 21, 2021?
Setler [38]

Answer:

No? Umm Hi Btw I Know You Might Now Give Me Brainliest But Can I Please Have Brainliest I Had Brainliest On My other accoun.t But The Next Day It Got hacke.d That's Why Im Asking I Kept Trying To Login To My Hack.ed Account Then A Thing That says Your account Got Haccked Shows Up Everytime But Now ITs NoT Saying My Account got hacke.d anymore It Just Wont Let Me in Now.

Explanation:

7 0
3 years ago
A 50g ball is released from rest 1.0 above the bottom of thetrack
ludmilkaskok [199]

Answer:

The maximum height of the ball is 2 m.

Explanation:

Given that,

Mass of ball = 50 g

Height = 1.0 m

Angle = 30°

The equation is

y=\dfrac{1}{4}x^2

We need to calculate the velocity

Using conservation of energy

\Delta U_{i}+\Delta K_{i}=\Delta K_{f}+\Delta U_{f}

Here, ball at rest so initial kinetic energy is zero and at the bottom the potential energy is zero

\Delta U_{i}=\Delta K_{f}

Put the value into the formula

mgh=\dfrac{1}{2}mv^2

Put the value into the formula

50\times10^{-3}\times9.8\times1.0=\dfrac{1}{2}\times50\times10^{-3}\times v^2

v^2=\dfrac{2\times50\times10^{-3}\times9.8\times1.0}{50\times10^{-3}}

v=\sqrt{19.6}

v=4.42\ m/s

We need to calculate the maximum height of the ball

Using again conservation of energy

\dfrac{1}{2}mv^2=mgh

Here, h = y highest point

Put the value into the formula

\dfrac{1}{2}\times50\times10^{-3}\times(4.42)^2=50\times10^{-3}\times9.8\times h

y=\dfrac{0.5\times(4.42)^2}{9.8}

y=0.996\ m

Put the value of y in the given equation

y=\dfrac{1}{4}x^2

x^2=4\times0.996

x=\sqrt{4\times0.996}

x=1.99\ m\ \approx 2 m

Hence, The maximum height of the ball is 2 m.

4 0
3 years ago
what is the efficiency of a machine that can carry a load of 100 kg with an effort of 20N when its velicity ratio is 10​ ASAP
Salsk061 [2.6K]

Answer:

Efficiency of machine=4.9

Explanation:

We are given that

Mass of load=100 kg

Effort=20 N

Velocity ratio=10

We have to find the efficiency of machine.

Load=mg=100\times 9.8=980 N

Efficiency=\frac{Load}{effort}\times \frac{1}{velocity\;ratio}

Using the formula

Efficiency of machine=\frac{980}{20}\times \frac{1}{10}

Efficiency of machine=\frac{49}{10}

Efficiency of machine=4.9

7 0
4 years ago
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