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Lorico [155]
3 years ago
13

Grade 11 question paper or memo for 2015

Physics
2 answers:
denis-greek [22]3 years ago
5 0
What your saying doesn't make sense.
stira [4]3 years ago
4 0

Answer:

wym?

sry i dont comprehend

Explanation:

You might be interested in
Viruses use the energy of the ______ cells to reproduce themselves​
galben [10]
<h2>Hey there!</h2>

<h2>The answer will be:</h2>

<h3>"Host"</h3>

<h2>Explanation:</h2><h2 /><h3>Viruses use the energy of the <u>host</u> cells to reproduce themselves​.</h3>

<h2>Hope it help you</h2>

5 0
3 years ago
A 56.0 kg bungee jumper jumps off a bridge and undergoes simple harmonic motion. If the period of oscillation is 11.2 s, what is
andre [41]

Answer:

2.80N/m

Explanation:

Given data

mass m= 56kg

perios T= 11.2s

The expression for the period is given as

T=2π√m/k

Substitute

11.2= 2*3.142*√56/k

square both sides

11.2^2= 2*3.142*56/k

125.44= 351.904/k

k=351.904/125.44

k= 2.80N/m

Hence the spring constant is 2.80N/m

8 0
2 years ago
An unstrained horizontal spring has a length of 0.26 m and a spring constant of 180 N/m. Two small charged objects are attached
MakcuM [25]

Answer:

a)

two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign. both charges are positive(+) or Negative (-)

b)

both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

Explanation:

Given that;

L = 0.26 m

k = 180 N/m

x = 0.039 m

a)

we know that two like charges always repel each other while two unlike charges attract each other. Since the spring stretches by 0.039 m, the charges have the same sign.

b)

Spring force F = kx

F = 180 × 0.039

F = 7.02 N

Now, Electrostatic force F = Keq²/r²

where r = L + x = ( 0.26 + 0.039 )

we know that proportionality constant in electrostatics equations Ke = 9×10⁹ kg⋅m3⋅s−2⋅C−2

so from the equation; F = Keq²/r²

Fr² = Keq²

q = √ ( Fr² / Ke )

we substitute

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q = √ ( 7.02 N × ( 0.26 + 0.039 )² / 9×10⁹ )

q =  √ (0.627595 / 9×10⁹)

q = √(6.97 × 10⁻¹¹)

q = 8.35 × 10⁻⁶ C

Therefore both q1 and q1 are 8.35 × 10⁻⁶ C or -8.35 × 10⁻⁶ C

5 0
2 years ago
the net external force on the 24-kg mower is stated to be 51 N. If the force of friction opposing the motion is 24 N, what force
-Dominant- [34]

Answer:

PART A)

External force will be 75 N

PART B)

distance moved will be 1.125 m

Explanation:

PART A)

Given that net force on the mower is

F_{net} = 51 N

now we also know that friction force due to ground is given as

F_f = 24 N

now we have

F_{net} = F_{ext} - F_f

51 = F_{ext} - 24

F_{ext} = 75 N

so external force will be 75 N

PART B)

deceleration due to friction when external force is removed from it

a = \frac{F_f}{m}

a = \frac{24}{24} = 1 m/s^2

now we can find the distance by kinematics

v_f^2 - v_i^2 = 2 a d

0 - 1.5^2 = 2(-1)d

d = 1.125 m

so the distance moved will be 1.125 m

6 0
3 years ago
How much work does a 60kg person do in walking up 5m of stairs
Nataliya [291]

Answer:

If I’m correct 300 joules

Explanation:

8 0
2 years ago
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