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blsea [12.9K]
3 years ago
8

A 30.0 kg object has an initial velocity of 7.00 m/s and a final velocity of 10.0 m/s. The work done on the object is __________

_ J. (Report the answer to three significant figures.)

Physics
2 answers:
kozerog [31]3 years ago
5 0

Answer:

765j

Explanation:

xz_007 [3.2K]3 years ago
3 0
Hope this helps :) If in need of clarification, feel free to ask!

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The best rebounders in basketball have a vertical leap (that is, the vertical movement of a fixed point on their body) of about
nadya68 [22]

Answer:

a) 4.45 m/s

b) 0.9 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -9.81\times 1}\\\Rightarrow u=4.45\ m/s

a) The vertical speed when the player leaves the ground is 4.45 m/s

v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-4.45}{-9.81}\\\Rightarrow t=0.45\ s

Time taken to reach the maximum height is 0.45 seconds

s=ut+\frac{1}{2}at^2\\\Rightarrow 1=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1\times 2}{9.81}}\\\Rightarrow t=0.45\ s

Time taken to reach the ground from the maximum height is 0.45 seconds

b) Time the player stayed in the air is 0.45+0.45 = 0.9 seconds

6 0
3 years ago
an ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uni
stiv31 [10]
The only vertical forces are weight and normal force, and they balance since the surface is horizontal. The horizontal forces are the applied force (uppercase F) in the direction the block slides and the frictional force (lowercase f) in the opposite direction.

Apply Newton's 2nd Law in the horizontal direction:
ΣF = ma
F - f = ma
where f = µmg

F - µmg = ma
F = m(a +µg)
F = (20 kg)(1.4 m/s² + 0.28(9.8 m/s²)

F = 83 N
3 0
3 years ago
what will be the moment of inertia of a body if it rotates at a uniform rate of 10rad/sec^2 by a torque of 120Nm?​
Naya [18.7K]

Answer:

12 kgm²

Explanation:

here angular acceleration = 10rad/sec²

torque= 120Nm

moment of inertia=?

we know,

torque= angular acceleration× moment of Inertia

or, moment of inertia = torque/angular acceleration

= 120/10

= 12kgm²

7 0
3 years ago
A magnetic field of magnitude 0.550 T is directed parallel to the plane of a circular loop of radius 43.0 cm. A current of 5.80
Assoli18 [71]

Answer:

\mu = 3.36\times 10^{-3}\ A-m^2

Explanation:

Given that,

The magnitude of magnetic field, B = 0.55 T

The radus of the loop, r = 43 cm = 0.43 m

The current in the loop, I = 5.8 mA = 0.0058 A

We need to find the magnetic moment of the loop. It is given by the relation as follows :

\mu = AI\\\\\mu=\pi r^2\times I

Put all the values,

\mu=\pi \times (0.43)^2\times 0.0058\\\\=3.36\times 10^{-3}\ A-m^2

So, the magnetic moment of the loop is equal to3.36\times 10^{-3}\ A-m^2.

3 0
3 years ago
Which of the following are appropriate acceleration units? km/hr2 m/s/s ft/s miles/hr/min sec/km/m
Reika [66]
I think it is m/s/s or miles/hr/min
8 0
4 years ago
Read 2 more answers
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