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Arlecino [84]
2 years ago
8

Helpp pleaseeeeee …..

Physics
1 answer:
Lilit [14]2 years ago
8 0

Answer:

300 cos 30 = 40 a + 40 * .2 * 10

Total force = mass * acceleration + frictional force

260 = 40 a + 80

a = 180 / 40 = 4.5 m/s^2

Check:

15 a + 15 * 10 * .2 = T    acceleration of 15 kg block (assuming a = 4.5)

T = 15 (4.5) + 30 = 97.5     force required to accelerate 15 kg block

260 - 97.5 = 162.5     net force on 25 kg block

162.5 = 4.5 (25) + 25 * 10 * .2

162.5 = 112.5 + 50 = 162.5

4.5 m/s^2 checks out as correct

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2kg + 3 kg = 5 kg as result of the conclusion
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3 years ago
A grape is thrown straight up in the air. If it was thrown at 2 m/s, how high will it go in the air?
ser-zykov [4K]

Answer:

Answer: <u>Height</u><u> </u><u>is</u><u> </u><u>0</u><u>.</u><u>2</u><u>0</u><u>4</u><u> </u><u>m</u>

Explanation:

At the highest point, it is called the maximum height.

• From third newton's equation of motion:

{ \rm{ {v}^{2}  =  {u}^{2}  + 2gs}}

• At maximum height, v is zero

• u is initial speed

• g is -9.8 m/s²

• s is the height

{ \rm{0 {}^{2}  =  {2}^{2}   -  (2 \times 9.8 \times s)}} \\  \\ { \rm{4 = 19.6s}} \\  \\ { \rm{s = 0.204 \: m}}

4 0
2 years ago
Read 2 more answers
The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the
Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t      ⇒     α = (ωfin - ωin) / t

⇒     α = (0 rev/min - 113 rev/min) / (60 min)

⇒     α = - 1.883 rev/min²

6 0
3 years ago
An 85 kg man and his 35 kg daughter are sitting on opposite ends of a 3.00 m see-saw. The see-saw is anchored in the center. If
wolverine [178]

Answer:

0.54m

Explanation:

Step one:

given data

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mass of man m1= 85kg

weight = mg

W1= 85*10= 850N

mass of daughter m2= 35kg

W2= 35*10= 350N

distance from the center= (1.5-0.2)= 1.3m

Step two:

we know that the sum of clockwise moment equals the anticlockwise moment

let the distance the must sit to balance the system be x

taking moment about the center of the system

350*1.3=850*x

455=850x

divide both sides by  850

x=455/850

x=0.54

Hence the man must sit 0.54m from the right to balance the system

3 0
3 years ago
When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin
mario62 [17]

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

4 0
3 years ago
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