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VashaNatasha [74]
3 years ago
12

A squirrel jumps down from a tall tree. Assuming the squirrel is in free fall, how far will it have fallen in 1.5 seconds?

Physics
2 answers:
Vlada [557]3 years ago
8 0

Answer:

d = 11.025 m

Explanation:

Since the situation is considered to be free fall so here given data as

initial speed = 0 m/s as it starts from rest

acceleration = 9.8 m/s/s

time of fall = 1.5 s

now we know from kinematics

d = v_i t + \frac{1}{2}at^2

d = 0 + \frac{1}{2}(9.8)(1.5)^2

d = 11.025m

so the distance traveled by the squirrel when it jumps from the tall tree is given as d = 11.025 m

snow_tiger [21]3 years ago
7 0
D = 1/2gt^2 <-- free fall formula

d = (0.5)(9.81)(1.5^2)
d = 11.03625
d = 11.04 m

hope this helps and have a great day :)




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Answer:

A) mass = 3121.58 kg

B) tension = 25940.37 N

C) tension = 25940.37 N (tension on both sides will be the same)

Explanation:

Weight of elevator = 22500 N

Distance = 6.75 m

Time = 3 sec

Since it started from rest, initial speed is zero.

Using Newton's equation of motion we have,

S = ut + 0.5at^2

S = distance covered = 6.75 m

t = time = 3 s

a = acceleration upwards

u = initial velocity = 0

Substituting values, we have,

6.75 = 0(3) + (0.5 x a x 3^2)

6.75 = 4.5a

a = 6.75/4.5 = 1.5 m/s^2 (acceleration of the elevator upwards)

Mass of the elevator = Weight/g

Where g = acceleration due to gravity 9.81 m/s

Mass = 22500/9.81 = 2293.58 kg

From the image below we solve from

T - 22500 = ma

T - 22500 = 2293.58 x 1.5

T - 22500 = 3440.37

T = 3440.37 + 22500 = 25940.37 N (this is the tension on the rope)

On the other side,

mg - T = ma

9.81m - 25940.37 = 1.5m

(9.81 - 1.5)m = 25940.37

8.31m = 25940.37

m = 3121.58 kg (mass of counter weight)

See image below

8 0
3 years ago
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