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valentina_108 [34]
3 years ago
10

Which is the mode of the following data?120, 125, 130, 125, 135​

Physics
2 answers:
Aleks [24]3 years ago
8 0

The mode in this case would be 125 because it occurs the most in the sequence of numbers.

miss Akunina [59]3 years ago
5 0

Answer:

125

Explanation:

mode is most occurring number

in this data 125 comes two times other numbers only come once hence mode is 125

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â/8.37 points scalcet8 12.2.037. ask your teacher my notes question part points submissions used a block-and-tackle pulley hoist
AnnyKZ [126]
Refer to the diagram shown below.

The hoist is in static equilibrium supported by tensions in the two ropes.

For horizontal force balance, obtain
T₃ cos 50 = T₂ cos 38
0.6428T₃ = 0.788T₂
T₃ = 1.2259T₂             (1)

For vertical force balance, obtain
T₂ sin 38 + T₃ sin 50 = 350
0.6157T₂ + 0.766T₃ = 350     (2)

Substitute (1) into (2).
0.6157T₂ + 0.766(1.2259T₂) = 350
1.5547T₂ = 350
T₂ = 225.124 N
T₃ = 1.2259(225.124) = 275.979

Answer:
T₂ = 225.12 N
T₃ = 275.98 N

4 0
3 years ago
How long does it take for a train to increase its velocity from 10m/s to 40m/s if it accelerates at 3 m/s
DIA [1.3K]

Answer:

Explanation:

Givens

Vi = 10 m/s

Vf = 40 m/s

a = 3 m/s^2

Formula

a = (vf - vi) /t              Substitute the givens into this formuls

Solution

3 = (40 - 10) / t          Multiply both sides by t

3*t = t(40 - 10)/t        Combine. Cancel t's on the right

3*t = 30                     Divide by 3

3t/3 = 30 / 3

Answer: t = 10 seconds.

6 0
2 years ago
What happens to the volume of a gas when pressure is applied? A. increases B. decreases C. Stays the same D. evaporates
BaLLatris [955]

Answer:

Decreases

Explanation:

8 0
3 years ago
A grocery cart with a mass of 15 kg is being pushed at constant speed up a 12∘ ramp by a force FP which acts at an angle of 17∘
Alenkasestr [34]

Answer: a. 198.6J b. - 198.6J

Explanation: Parameters given:

m = 15kg

g = 9.8m/s²

∅ = 12°

a. Work done by the force Fp on the cart if the ramp is 6.5m long.

Given the formula, Fp = Mgsin∅ = 15 x 9.8 x sin12° = 30.56N

Therefore Work done (Wp) = Fp x Ramp Length = 30.56 x 6.5 = 198.64Nm or 198.6J

b. The work done by the force mg on the cart.

Since the cart is being pushed upwards, it acts against gravity with its direction of motion. Taking into account the formula from the previous answer for Work Done (Wg) = Fmg x distance

= 15kg x -9.8m/s² x Sin12° x 6.5m

= - 198.6J

4 0
3 years ago
Read 2 more answers
julia throws a ball vertically upward from the ground with a speed of 5.89m/s. Andrew catches it when it is on its way down at a
aliya0001 [1]
Vo = 5.89 m/s Y = 1.27 m g = 9.81 m/s^2 
Time to height 
Tr = Vo / g Tr = (5.89 m/s) / (9.81 m/s^2) Tr = 0.60 s 
Max height achieved is:
H = Vo^2 / [2g] H = (5.89 )^2 / [ 2 * (9.81) ] H = (34.69) / [19.62] H = 1.77 m 
It falls that distance, minus Andrew's catch distance:
h = H - Y h = (1.77 m) - (1.27 m) h = 0.5 m 
Time to descend is therefore:
Tf = √ { [2h] / g ] Tf = √ { [ 2 * (0.5 m) ] / (9.81 m/s^2) } Tf = √ { [ 1.0 m ] / (9.81 m/s^2) } Tf = √ { 0.102 s^2 } Tf = 0.32 s 
Total time is rise plus fall therefore:
Tt = Tr + Tf Tt = (0.60 s) + (0.32 s) Tt = 0.92 s           (ANSWER)
8 0
3 years ago
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