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2 years ago

Example of extensive property Please fast

Physics

1 answer:

34kurt2 years ago

3 0

Answer:

Mass and volume

Explanation:

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Help me, i don't know the answer

8_murik_8 [283]

Explanation:

A) Bb BB

B) 50%

A) 50%

B) b. b

B. Bb. Bb

b. bb. bb

4 0

2 years ago

Is it using energy when someone stands against a locker that doesn’t move

soldi70 [24.7K]

When someone stands against a locker and is does not moving at all, then there will be no displacement and since displacement = 0

Work done also becomes equal to zero.

Work done is usually defined as change in energy. Since the work done is zero there has been no energy used.

5 0

2 years ago

A wagon full of manure accidentally rolls down a driveway for 5.0m while a person pushes against the wagon with a force of 420 N

Cerrena [4.2K]

Answer:

2100 J

Explanation:

Parameters given:

Force acting on the object, F = 420 N

Distance moved by object, d = 5m

The change in kinetic energy of an object is equal to the work done by a force acting on the object:

W = F * d

∆KE = F * d

∆KE = 420 * 5

∆KE = 2100 J

8 0

3 years ago

Read 2 more answers

A disk-shaped merry-go-round of radius 2.63 m and mass 152 kg rotates freely with an angular speed of 0.526 rev/s. A 51.7 kg per

zalisa [80]

Explanation:

Given

radius $r=2.63 m$

$N=0.526 rev/s$

$\omega =3.30 rad/s$

mass disc $M=152 kg$

mass of person $m=51.7 kg$

velocity of Person $v=2.76 m/s$

moment of inertia $I=Mr^2$

$I=0.5\times 152\times 2.63^2=827.64 kg-m^2$

Initial angular momentum

$L_i=I\omega +mvr$

$L_i=827.64\times 3.30+51.7\times 2.76\times 2.63$

$L_i=2731.212+375.27=3106.48 Js$

Final Moment inertia

$I_f=0.5Mr^2+mr^2$

$I_f=(152\cdot 0.5+51.7)\cdot (2.63)^2=1185.243 kg-m^2$

final angular momentum

$L_f=I_f\omega _f$

Conserving angular momentum

$L_i=L_f$

$3106.48=1408.97\times \omega _f$

$\omega _f=2.62 rad/s$

4 0

2 years ago

A block with mass m1 = 8.8 kg is on an incline with an angle θ = 41.0° with respect to the horizontal. For the first question th

serg [7]

So based on your question where there is a block of mass m1= 8.8kg in the inclined plane with an angle of 41 with respect to the horizontal. To find the spring constant of the problem were their is a coefficients of friction of 0.39 and 0.429, you must use the formula K*x^2=m*a*sin(angle). By calculating the minimum spring constant is 220.66 N/m^2

6 0

3 years ago