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Tomtit [17]
3 years ago
5

In object of mass 30 kg is in freefall in a vacuum where there is no air resistance determine the acceleration of the object

Physics
1 answer:
Leto [7]3 years ago
3 0
The free fall acceleration due to the Earth's gravity will always be 9.8m/s^2. So, if this question is taking place on Earth, then the acceleration of an object in free fall is 9.8m/^2.
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6 0
3 years ago
A 280-m-wide river flows due east at a uniform speed of 4.7m/s. A boat with a speed of 7.1m/s relative to the water leaves the s
tamaranim1 [39]

Answer:

(a) The speed is 7.96 m/s

(b) The direction is 76 degree from positive X axis in counter clockwise direction.  

Explanation:

Width of river = 280 m

speed of river, vR = 4.7 m/s towards east

speed of boat with respect to water, v(B,R) = 7.1 m/s at 26 degree west of north

vR = 4.7 i \\\\v(B,R) = 7.1 (- sin 26 i + cos 26 j) = - 3.1 i + 6.4 j

(a) The velocity of boat with respect to ground is

\overrightarrow{v}_{(B,R)}=\overrightarrow{v}_{(B,G)}-\overrightarrow{v}_{(R,G)}\\\\- 3.1 \widehat{i} +6.4 \widehat{j}=\overrightarrow{v}_{(B,G)} - 4.7 \widehat{i}\\\\\overrightarrow{v}_{(B,G)} = 1.6 \widehat{i} + 6.4 \widehat{j}\\\\{v}_{(B,G)} = \sqrt{1.6^2 + 6.4^2}=6.96 m/s

(b) The direction is given  by

tan\theta = \frac{6.4}{1.6} =4\\\\\theta = 76^o

7 0
2 years ago
Two trains on separate tracks move toward each other. Train 1 has a speed of 145 km/h; train 2, a speed of 72.0 km/h. Train 2 bl
tekilochka [14]

Answer:

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

Explanation:

Given:

Two trains on separate tracks move toward each other

For Train 1 Velocity of the observer,

v_{o}=145\ km/h=145\times \dfrac{1000}{3600}=40.28\ m/s

For Train 2 Velocity of the Source,

v_{s}=90\ km/h=90\times \dfrac{1000}{3600}=25\ m/s

Frequency of Source,

f_{s}=500\ Hz

To Find:

Frequency of Observer,

f_{o}=?  (frequency heard by the engineer on train 1)

Solution:

Here we can use the Doppler effect equation to calculate both the velocity of the source v_{s} and observer v_{o}, the original frequency of the sound waves f_{s} and the observed frequency of the sound waves f_{o},

The Equation is

f_{o}=f_{s}(\dfrac{v+v_{o}}{v -v_{s}})

Where,

v = velocity of sound in air = 343 m/s

Substituting the values we get

f_{o}=500(\dfrac{343+40.28}{343 -25})=500\times 1.205=602.64\approx 603\ Hz

Therefore,

The frequency heard by the engineer on train 1

f_{o}=603\ Hz

7 0
3 years ago
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