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elena-14-01-66 [18.8K]

2 years ago

11

gabriel and arely have performed an experiment and are not sure their results are valid. what should they do to check their resu

lts

2 answers:

zubka84 [21]2 years ago

7 0

Do the math and then compare your results to theirs.

Arisa [49]2 years ago

6 0

Yeah, or they can compare the results to other people

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A man weighs himself twice in an elevator. When the elevator is at rest, he weighs 824 N; when the elevator starts moving upward

kifflom [539]

Answer: c. 1.3 m/s^2

Explanation:

When he is at rest, is weight can be calculated as:

W = g*m

where:

m = mass of the man

g = gravitational acceleration = 9.8m/s^2

We know that at rest his weight is W = 824N, then we have:

824N = m*9.8m/s^2

824N/(9.8m/s^2) = m = 84.1 kg

Now, when the elevators moves up with an acceleration a, the acceleration that the man inside fells down is g + a.

Then the new weight is calculated as:

W = m*(g + a)

and we know that in this case:

W = 932N

g = 9.8m/s^2

m = 84.1 kg

Then we can find the value of a if we solve:

932N = 84.1kg*(9.8m/s^2 + a)

932N/84.1kg = 11.1 m/s^2 = 9.8m/s^2 + a

11.1 m/s^2 - 9.8m/s^2 = a = 1.3 m/s^2

The correct option is C

3 0

2 years ago

A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad

dalvyx [7]

The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s

3 0

3 years ago

A helicopter has blades of length 4.0 m rotating at 3.0 rev/s in a horizontal plane.If the vertical component of the Earth’s mag

VARVARA [1.3K]

Answer:

Induced emf, $\epsilon=9.79\times 10^7\ volts$

Explanation:

Given that,

Length of the helicopter, l = 4 m

Angular speed of the helicopter, $\omega=3\ rev/s=18.84\ rad/s$

The vertical component of the Earth’s magnetic field is, $B=6.5\times 10^5\ T$

We need to find the induced emf between the tip of a blade and the hub. The induced emf in terms of angular velocity of an rotating object is given by :

$\epsilon=\dfrac{1}{2}B\omega l^2$

$\epsilon=\dfrac{1}{2}\times 6.5\times 10^5\times 18.84\times (4)^2$

$\epsilon=9.79\times 10^7\ volts$

So, the induced emf between the tip of a blade and the hub is $\epsilon=9.79\times 10^7\ volts$ . Hence, this is the required solution.

5 0

2 years ago

You push a cart with mass 15 kg forward , giving it an acceleration of 3 m/s ^ 2 . How much force did you apply ? O A. 0.2N O B.

Alexandra [31]

Answer: force = mass x accelegation

Explanation: 15x3 = 45 N

7 0

2 years ago

Read 2 more answers

A hockey puck has a mass of 0.107 kg and is at rest. A hockey player makes a shot, exerting a constant force of 28.0 N on the pu

gregori [183]

Answer:

The speed does it head toward the goal = 41.87 $\frac{m}{s}$

Explanation:

Mass = 0.107 kg

Initial velocity ( u ) = 0

Force (F) = 28 N

Time = 0.16 sec

From newton's second law, Force = mass × acceleration

⇒ F = m × a

⇒ 28 = 0.107 × a

⇒ a = 261.7 $\frac{m}{s^{2} }$ --------- (1)

This is the value of acceleration.

Final speed of the mass is calculated by the equation V = U + at

⇒ U = 0 because mass in in rest position at start.

⇒ V = a t

Put the values of acceleration and time in above formula we get

⇒ V = 261.7 × 0.16

⇒ V = 41.87 $\frac{m}{s}$

Therefore the speed does it head toward the goal = 41.87 $\frac{m}{s}$

5 0

3 years ago