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artcher [175]
3 years ago
8

The slope of the following graph represents

Physics
1 answer:
guajiro [1.7K]3 years ago
4 0

Answer:

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A uniform rod of length 0.8 m and mass 1.4 kg, has two point masses at each end. The point mass on the left end has a mass 1.2 k
VladimirAG [237]

Answer:

Explanation:

1.2(0) + 3(0.8) + 1.4(0.8/2) / (1.2 + 3 + 1.4) = 0.5285714... ≈ 0.53 m

5 0
2 years ago
A potential difference of 24 V is applied to a 150-ohm resistor. How much current flows through the resistor?​
Lady_Fox [76]

Given :- A resistor of 150 ohm, hence Resistance (R) = 150 ohm

Potential Difference (v) = 24 V

Current (I) = ?

V = IR

24 = I × 150

I = 24/150

I = 0.16 ampere

hope it helps!

3 0
2 years ago
To demonstrate the tremendous acceleration of a top fuel dragracer, you attempt to run your car into the back of a dragster that
noname [10]

Answer:

a. 2v₀/a   b. 2v₀/a  

Explanation:

a. Since you are moving with a constant velocity v₀, the distance, s you cover in time = t max is s = v₀t.

Since the dragster starts from rest with an acceleration, a, using

s' = ut + 1/2at² where u = 0 and s' = distance moved by dragster

s' = 0t + 1/2at²

s' = 1/2at²

Since the distance moved by me and the dragster must be the same,

s = s'

v₀t. =  1/2at²

v₀t. - 1/2at² = 0

t(v₀ - 1/2at) = 0

t= 0 or v₀ - 1/2at = 0

t= 0 or v₀ = 1/2at

t= 0 or t = 2v₀/a  

So the maximum time tmax = 2v₀/a

b. Since the distance covered by me to meet the dragster is s = v₀t in time, t = tmax which is also my distance from the dragster when it started. So, my distance from the dragster when it started is s =  v₀(2v₀/a)

= 2v₀/a  

4 0
3 years ago
A bowling (mass = 7.2 kg, radius = 0.11 m) and a billiard ball (mass = 0.38 kg, radius = 0.028 m) may each be treated as uniform
cestrela7 [59]
Hope this helps you!

7 0
2 years ago
20. A mass of gas has a volume of 4 m3, a temperature of 290 K, and an absolute pressure of 475 kPa. When the gas is allowed to
aliina [53]
Given:\\V_1=4m^3\\T_1=290K\\p_1=475kPa\\\\V_2=6.5m^3\\T_2=277K\\\\Find:\\p_2=?\\\\Solution:\\\\ \frac{pV}{T} =const.\\\\ \frac{p_1V_1}{T_1} = \frac{p_2V_2}{T_2} \\\\\frac{p_1V_1T_2}{T_1}=p_2V_2\\\\\frac{p_1V_1T_2}{T_1V_2}=p_2\\\\p_2=p_1 \frac{V_1}{V_2}  \frac{T_2}{T_1} \\\\\\p_2=475kPa\cdot  \frac{4m^3}{6.5m^3} \cdot  \frac{277K}{290K} \approx 279.2kPa\\\\Correct\;is\;answer\;\;C
7 0
2 years ago
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