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Vinil7 [7]
3 years ago
6

Need help on question one ASAP pls help

Physics
1 answer:
Anon25 [30]3 years ago
6 0

Answer:

B

Explanation:

Because blocks 2 and 3 have sides with unequal forces while block 1 doesnt. Plz drop a follow if this helps! ❤

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A spring has a force constant k, and an object of mass m is suspended from it. The spring is cut in half and the same object is
kenny6666 [7]

Answer:

f2/f1 = \sqrt{2}

Explanation:

From frequency of oscillation

f = 1/2pi *\sqrt{k/m}

Initially with the suspended string, the above equation is correct for the relation, hence

f1 = 1/2pi *\sqrt{k/m}

where k is force constant and m is the mass

When the spring is cut into half, by physics, the force constant will be doubled as they are inversely proportional

f2 = 1/2pi *\sqrt{2k/m}

Employing f2/ f1, we have

f2/f1 = \sqrt{2}

3 0
3 years ago
A nail driven into a board increases in temperature.
denis23 [38]

Answer:

ΔT = 40.91 °C

Explanation:

First we find the kinetic energy of one hit to the nail:

K.E = (1/2)mv²

where,

K.E = Kinetic energy = ?

m = mass of hammer = 1.6 kg

v = speed of hammer = 7.7 m/s

Therefore,

K.E = (1/2)(1.6 kg)(7.7 m/s)²

K.E = 47.432 J

Now, for 10 hits:

K.E = (10)(47.432 J)

K.E = 474.32 J

Now, we calculate the heat energy transferred (Q) to the nail. As, it is the 59% of K.E. Therefore,

Q = (0.59)K.E

Q = (0.59)(474.32 J)

Q = 279.84 J

The change in energy of nail is given as:

Q = mCΔT

where,

m = mass of nail = 7.6 g = 0.0076 kg

C = specific heat capacity of aluminum = 900 J/kg.°C

ΔT = Increase in temperature = ?

Therefore,

279.84 J = (0.0076 kg)(900 J/kg.°C)ΔT

ΔT = (279.84 J)/(6.84 J/°C)

<u>ΔT = 40.91 °C</u>

5 0
3 years ago
Car A m= 2000kg<br> v=20 km/h <br> Car B m= 2000kg <br> v= 20km/h what is the location after impact
Darya [45]

Mass of car, M=2000kg

Velocity, V=72× 5/18 =20ms¹

Apply kinematic equation of motion

v² -u² = as

0 − 20² =2a×20

a=−10ms^-2

Breaking force, F=ma=2000×10=20 kN

Apply first kinematic equation

v = u +at

t= \frac{v - u}{a}  =  \frac{0 - 20}{ - 10}  = 2sec\\

7 0
3 years ago
Help Please? Hshshahahahshshhs
pickupchik [31]

the answer is no change

7 0
4 years ago
A spring with a spring constant of 25.0N/m is stretched 4.50m. What is the force that the spring would apply?​
alina1380 [7]

Explanation:

from \: hookes \: law : \\ force = k \bold{.}e \\  = 25.0 \times 4.50 \\  = 112.5 \: N

8 0
3 years ago
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