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tresset_1 [31]
3 years ago
11

the magnitude of the normal force acting on a person with mass of 70 kg standing at rest on the flat ground would be ?

Physics
1 answer:
Mekhanik [1.2K]3 years ago
3 0

Answer:

f = mg \\  = 70 \times 9.8 =  |f|

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a 2000kg car initially traveling at a speed of 15 m/s is accelerated by a constant force of 10000 n for 3 seconds. the new speed
Juliette [100K]
  • Mass of the car (m) = 2000 Kg
  • Initial velocity (u) = 15 m/s
  • Force (F) = 10000 N
  • Time (t) = 3 s
  • Let the acceleration be a.
  • By using the formula, F = ma, we get,
  • 10000 N = 2000 Kg × a
  • or, a = 10000 N ÷ 2000 Kg
  • or, a = 5 m/s^2
  • Let the final velocity be v.
  • By using the formula, v = u + at, we get,
  • v = 15 m/s + 5 m/s^2 × 3 s
  • or, v = 15 m/s + 15 m/s
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<u>Answer</u><u>:</u>

<em><u>The </u></em><em><u>new </u></em><em><u>sp</u></em><em><u>e</u></em><em><u>ed </u></em><em><u>of </u></em><em><u>the </u></em><em><u>car </u></em><em><u>is </u></em><em><u>3</u></em><em><u>0</u></em><em><u> </u></em><em><u>m/</u></em><em><u>s.</u></em>

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8 0
3 years ago
Breanna is standing beside a merry-go-round pushing 19° from the tangential direction and is able to accelerate the ride and her
leva [86]

To solve the problem it is necessary to apply the concepts given in the kinematic equations of angular motion that include force, acceleration and work.

Torque in a body is defined as,

\tau_l = F*d

And in angular movement like

\tau_a = I*\alpha

Where,

F= Force

d= Distance

I = Inertia

\alpha = Acceleration Angular

PART A) For the given case we have the torque we have it in component mode, so the component in the X axis is the net for the calculation.

\tau= F*cos(19)*d

On the other hand we have the speed data expressed in RPM, as well

\omega_f = 10rpm = 10\frac{1rev}{1min}(\frac{1min}{60s})(\frac{2\pi rad}{1rev})

\omega_f = 1.0471rad/s

Acceleration can be calculated by

\alpha = \frac{\omega_f}{t}

\alpha = \frac{1.0471}{9}

\alpha = 0.11rad/s^2

In the case of Inertia we know that it is equivalent to

I = \frac{1}{2}mr^2 = \frac{1}{2}(750)*(2.3)^2

I = 1983.75kg.m^2

Matching the two types of torque we have to,

\tau_l=\tau_a

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Fcos(19)*2.3=1983.75(0.11)

F=100.34N

PART B) The work performed would be calculated from the relationship between angular velocity and moment of inertia, that is,

W = \frac{1}{2}I\omega_f^2

W= \frac{1}{2}(1983.75)(1.0471)^2

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7 0
3 years ago
When two sticks are laid end-to-end they cover a length of 8.32 feet. One stick is 2.93 ft longer than the other. What is the le
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To solve this problem we will start by defining the length of the shortest stick as 'x'. And the magnitude of the longest stick, according to the statement as

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Now solving for x we have,

x + (x + 2.93) = 8.32

2x + 2.93 = 8.32

2x = 8.32 - 2.93

x = \frac{ 8.32 - 2.93}{2}

x = 2.695 ft

Therefore the shorter stick is 2.695ft long.

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Answer: 1. Carbon dioxide is not an air pollutant

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2 years ago
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Answer: to only change one factor in an experiment or test

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