Một vận động viên ném tạ quay một cái đĩa nặng 1kg với bán

A What quantity is equal to

natulia [17]

Answer:

Option B is correct

Explanation:

acceleration is equal to rate of change of velocity.

6 0

2 years ago

How much work should be done to lift a 5kg brick to the height of 12m

Goshia [24]

Answer: 588 joules

Explanation:

Work is done when force is applied on an object over a distance ( whether vertical or horizontal). It is measured in joules.

Thus, Workdone = Force X distance

- Vertical distance to be moved by the brick = 12 metres

- Mass of box = 5kg

- Acceleration due to gravity when box was lifted represented by g is a constant with value of 9.8m/s^2

Now, recall that Force = Mass x acceleration due to gravity

i.e Force = 5kg x 9.8m/s^2

Force = 49 Newton

So, Workdone = Force X Distance

Workdone = 49 Newton X 12 metres

Workdone = 588 joules

Thus, 588 joules of work was done.

8 0

2 years ago

Read 2 more answers

We calculated the speed of visible light through four substances. What is different about the substances? Lesson: 4.10 Substance

marusya05 [52]

Answer:

it's to dense

Explanation:

3 0

2 years ago

To practice Problem-Solving Strategy 27.1: Magnetic Forces. A particle with mass 1.81×10−3 kgkg and a charge of 1.22×10−8 CC has

Roman55 [17]

Answer:

The magnitude of the acceleration is $a = 0.33 m/s^2$

The direction is $- \r k$ i.e the negative direction of the z-axis

Explanation:

From the question we are that

The mass of the particle $m = 1.8*10^{-3} kg$

The charge on the particle is $q = 1.22*10^{-8}C$

The velocity is $\= v = (3.0*10^4 m/s ) j$

The the magnetic field is $\= B = (1.63T)\r i + (0.980T) \r j$

The charge experienced a force which is mathematically represented as

$F = q (\= v * \= B)$

Substituting value

$F = 1.22*10^{-8} (( 3*10^4 ) \r j \ \ X \ \ ( 1.63 \r i + 0.980 \r j )T)$

$= 1.22 *10^{-8} ((3*10^4 * 1.63)(\r j \ \ X \ \ \r i) + (3*10^4 * 0.980) (\r j \ \ X \ \ \r j))$

$= (-5.966*10^4 N) \r k$

Note :

$i \ \ X \ \ j = k \\\\j \ \ X \ \ k = i\\\\k \ \ X \ \ i = j\\\\j \ \ X \ \ i = -k \\\\k \ \ X \ \ j = -i\\\\i \ \ X \ \ k = - j\\$

Now force is also mathematically represented as

$F = ma$

Making a the subject

$a = \frac{F}{m}$

Substituting values

$a =\frac{(-5.966*10^4) \r k}{1.81*10^{-3}}$

$= (-0.33m/s^2)\r k$

$= 0.33m/s^2 * (- \r k)$

6 0

3 years ago

At what altitude above the earth's surface would the acceleration due to gravity be 4.9 m/s^2? Assume the mean radius of the ear

Mandarinka [93]

We apply the gravity calculation expressed in the formula: g=GM/r2
where G is the gravitational constant, m is the mass and r is the radius
r=√GM/g
(1) Radius = √6.674e-11*5.972e24/8 = 7058 kms Earth radius or surface of earth from center of earth= 6400 kmsSo r= 658 kms from surface of earth.
Gravity 8m/s2 will be at 658 kms from surface of earth.
(2) half gravity= 9.8/2= 4.9 m/s2 Radius=√6.674e-11*5.972e24/4.9 = 9019 kms Half Gravity will exist at 9019-6400= 2619 kms from surface of earth.

4 0

2 years ago