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3 years ago

Find the speed of a bicyclist who took an hour and a half to travel 10km

Physics

1 answer:

Natalija [7]3 years ago

7 0

distance = 10km

time = 1 1/2 = 3/2 hours

speed = distance/time = 10 / 3/2 = 20/3 = 6.67 Km/h

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A 50N girl pushes a 10,000 N car with force of 200N. What is the force the car pushes back at the girl? *

miss Akunina [59]

Answer:

200N

Explanation:

EVERY FORCE IS OPPOSED BY AN EQUAL FORCE, REGARDLESS OF THE WEIGHT OF THE OBJECTS APPLYING THE FORCE.

8 0

3 years ago

The coefficient of static friction between hard rubber and normal street pavement is about 0.90. On how steep a hill (maximum an

jok3333 [9.3K]

First thing to do is to draw the system described above. Then, write an equation for the forces present.

ΣF = Fg - Ff
0 = mgsin∅ - umgcos∅0 = gsin∅ - ugcos∅
u = tan∅
∅(max) = tan^-1 (u)

7 0

3 years ago

Read 2 more answers

Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while

Vinvika [58]

Answer:

Explanation:

Given

Time taken to reach ground is $t=2.7\ s$

Malda initial velocity $u=95\ cm/s$

Let h be the height of Cliff

using $h=ut+\frac{1}{2}at^2$

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

$h=0+\frac{1}{2}(9.8)(2.7)^2$

$h=35.72\ m$

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

$R=u\times t$

$R=0.95\times 2.7=2.43\ m$

7 0

3 years ago

Hydraulic engineers often use, as a unit of volume of water, the "acre-foot", defined as the volume of water that will cover 1 a

Alex17521 [72]

Answer:

Volume = 1,015 acre-feet (Approx)

Explanation:

Given:

Rain = 1.7 in

Time = 30 min

Area = 29 km²

Find:

Volume in acre-feet

Computation:

1 km = 1,000 m

1 m = 3.28 feet

1 km² = 247.105 acre

d = 1.7 in = 1.7 / 12 = 0.14167 ft

Area = 29 × 247.105 = 7,166.045 acre

Volume = 7,166.045 acre × 0.14167 ft

Volume = 1,015 acre-feet (Approx)

7 0

3 years ago

The flywheel of an engine has moment of inertia 2.50 kg m2 about its rotation axis. What constant torque is required to bring it

MrRissso [65]

Answer:

Explanation:

From the question we are told that

The moment of inertia is $I = 2.50 \ kg \cdot m^2$

The final angular speed is $w_f = 400 rev/min = \frac{400 * 2\pi}{60} = 41.89 \ rad/s$

The time taken is $t = 8.0 s$

The initial angular speed is $w_i = 0\ rad/s$

Generally the average angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{41.89}{8}$

=> $\alpha = 5.24 \ rad/s^2$

Generally the torque is mathematically represented as

$\tau = I * \alpha$

=> $\tau = 5.24 * 2.50$

=> $\tau = 13.09 \ N \cdot m$

5 0

3 years ago