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3 years ago

Recalculate 100 km/h to m/s

Physics

1 answer:

qwelly [4]3 years ago

4 0

= 27.777

Explanation:

A kilometer has 1,000 meters, and an hour has 3,600 seconds, so 100 kilometers per hour is: 100 x 1,000 / 3,600 = 27.777... m/s.

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The breaking in (blank) bonds in food releases energy for your body to use

kicyunya [14]

Answer:

The breaking in <em>molecular</em> bonds in food releases energy for your body to use.

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2 years ago

A horizontal uniform bar of mass 2.7 kg and length 3.0 m is hung horizontally on two vertical strings. String 1 is attached to t

Jlenok [28]

Answer:

14.36 N

Explanation:

$T_{1}$ = Tension in string 1

$T_{2}$ = Tension in string 2

$m_b$ = mass of the bar = 2.7 kg

$W_b$ = weight of the bar

weight of the bar is given as

$W_b = m_{b} g = (2.7) (9.8) = 26.46$ N

$m_m$ = mass of the bar = 1.35 kg

$W_m$ = weight of the monkey

weight of the monkey is given as

$W_m = m_{m} g = (1.35) (9.8) = 13.23$ N

Using equilibrium of torque about left end

$W_{m} (AB) + W_{b} (AB) = T_{2} (AC)\\W_{m} (AB) + W_{b} (AB) = T_{2} (AD - CD)\\(13.23) (1.5) + (26.46)(1.5) = T_{2} (3 - 0.65)\\\\T_{2} = 25.33$ N

Using equilibrium of force in vertical direction

$T_{1} + T_{2} = W_{b} + W_{m}\\T_{1} + 25.33 = 26.46 + 13.23\\T_{1} = 14.36$ N

7 0

3 years ago

Help pls they are all going to have the same answers.

Liula [17]

1.activities of daily living.
2.activity of daily living.
3.planned exercises activities.
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3 years ago

Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 10

Tema [17]

Answer:

Explanation:

(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.

Hence, his average velocity is 300m/150s=2ms^−1

(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.

Therefore, his average speed for this journey is 400m210s=1.9ms−1.

For the same journey is displacement is equal to the distance between the points A and C,i.e. 300m−100m=200m.

Hence, his average velocity for this case is 200m/210s=0.95ms^−1

7 0

3 years ago

A mosquito can fly with a speed of 1.10 m/s with respect to the air. Suppose a mosquito flies east at this speed across a swamp.

sineoko [7]

Answer:

The velocity of Mosquito with respect to earth will be 0.302m/s

Explanation:

V(ma) = 1.10 m/s, east Velocity of mosquito with respect to air

V(ae) = 1.4 m/s at 35° Velocity of air with respect to Earth in west of south direction.

Velocity of Mosquito with respect to earth will be

V(me) = V(ma) + V(ae)

We need to find the mosquito’s speed with respect to Earth in the x direction.

V(x, me) = V(x, ma) + V(x, ae) = V(ma) + V(ae)(cos theta(ae) )

Angle (ae) = –90.0° − 35°=−125°

V(x, me) = 1.10 + (1.4)Cos(-125)

= 1.10 + 1.4(-0.57)

= 1.10 -0.798

= 0.302

So the velocity of Mosquito with respect to earth will be 0.302m/s

7 0

3 years ago