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11Alexandr11 [23.1K]
3 years ago
10

A bullet of mass 120g is fired horizontally into a fixed wooden block with a speed of 20m\s. The bullet is brought to rest in a

wooden block in 0.1secs by a constant resistance
calculate the;(I) magnitude of the resistance
(ii) distance moved by the bullet in the wood​
Physics
1 answer:
prohojiy [21]3 years ago
4 0

Answer:

1) F = 24 N

2) Distance = 1 m

Explanation:

We are given;

Mass; m = 120 g = 0.12 kg

Initial velocity; u = 20 m/s

Final velocity; v = 0 m/s since it came to rest.

Time; t = 0.1 s

We can calculate acceleration from Newton's first equation of motion;

a = (v - u)/t

a = (0 - 20)/0.1

a = -200 m/s²

1) magnitude of the resistance will be;

F = ma

F = 0.12 × (-200)

F = -24 N

Since, we are dealing with the magnitude, we will take the absolute value. Thus, F = 24 N

2) To find the distance moved by the bullet, we know that;

Distance = Average speed × time

Thus;

Distance = ((v + u)/2) × t

Distance = ((0 + 20)/2) × 0.1

Distance = 1 m

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Answer:

204kW

Explanation:

P = t x ω = 500 x (2πx3900/60) = 204203W = 204kW

3 0
3 years ago
How much heat is released when 35kg of water freezes?
inn [45]
You need an additional point of data here: the enthalpy of fusion, or conversely the enthalpy of melting (they differ only by their sign). For water (or ice) that value is gotten from sources such as the internet

<span>ΔH°(fus) = 6.01 kJ/mole </span>

<span>Since you have 35 000g, how many moles do you have? </span>
<span>Moles H2O = 35000 g/(18.015 g/mole) = 1942.8 moles</span>

<span>So, take that ΔH°(fus) in kJ/mole, multiply by the number of moles, and there ya go! 
</span>
6.01 x 1942.8 = 11,676 kJ of energy is released

Hope I helped!! xx
3 0
3 years ago
A 1200-kg car is travelling east at a rate of 9 m/s. A 1600-kg truck is travelling south at a rate of 13 m/s. The truck accident
weeeeeb [17]

Answer:

Around 3.57m/s

Explanation:

p=mv

Let's denote the momentum, mass, and velocity of the car with the subscript 1, and for the truck use 2. After the collision, the combined momentum can be denoted with the subscript 3.

p_1=1200\cdot 9=10800 \\\\p_2=1600\cdot 13=20800 \\\\20800-10800=10000 \\\\1200kg+1600kg=2800kg \\\\10000=2800v_3 \\\\v_3\approx 3.57m/s

Hope this helps!

3 0
3 years ago
A transverse wave on a rope is given by y(x,t)= (0.750cm)cos(π[(0.400cm−1)x+(250s−1)t]). part a part complete find the amplitude
Pani-rosa [81]
The amplitude of a wave corresponds to its maximum oscillation of the wave itself. 
In our problem, the equation of the wave is
y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
y(x,t)=0.750 cm
and therefore this value corresponds to the amplitude of the wave.
4 0
3 years ago
Read 2 more answers
Calculate the energy released by the electron-capture decay of 5727Co. Consider only the energy of the nuclei (ignore the energy
erma4kov [3.2K]

Answer:

Explanation:

⁵⁷Co₂₇  + e⁻¹  =  ²⁷Fe₂₆

mass defect = 56.936296 + .00055 - 56.935399

= .001447 u

equivalent energy

= 931.5 x .001447 MeV

= 1.3479 MeV .

= 1.35 MeV

energy of gamma ray photons = .14  + .017

= .157 MeV .

Rest of the energy goes to neutrino .

energy going to neutrino .

= 1.35 - .157

= 1.193 MeV.

5 0
3 years ago
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