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laila [671]
3 years ago
6

Skateboard falls with force of weight pulling downwards with 53 Newton’s accelerating at a rate of 9.81m/s what’s the mass of th

e skateboard
Physics
1 answer:
Veseljchak [2.6K]3 years ago
8 0

Point of correction, acceleration is 9.81 m/s2

Answer:

5.4 Kg

Explanation:

We know that F=ma and making m the subject of the formula then

m=\frac {F}{a} where m is the mass of the skateboard, F is the pulling force and a is the acceleration of the body

Substituting 53 N for F and 9.81 m/s2 for a then

m=\frac {53 N}{9.81 m/s2}=5.40265 Kg\approx 5.4 Kg

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If a driver is tired, the thinking distance will be less. True or false.why?
Evgesh-ka [11]

Answer:

False

Explanation:

When a driver is impaired the brain will need an extra second or two to wake up, after it does wake up it tends to drift off regularly.

8 0
3 years ago
A playground slide is inclined 40°. If a boy with a mass of 32 kg slides down for -3meters. How much work is done by gravity on
telo118 [61]

Answer:

the work done by gravity on the boy is 604.62 J

Explanation:

Given;

distance the boy slides, d = 3 m

angle of inclination of the playground, θ = 40⁰

mass of the boy, m = 32 kg

The vertical height, h, above the ground through which the boy falls represents the height of the triangle which is the opposite side.

The distance through which the boy slides, d, represents the hypotenuse side of the right triangle.

sin \theta = \frac{opposite }{hypotenuse} \\\\sin \theta = \frac{h}{d} \\\\h = dsin\theta\\\\h = 3 \times sin(40^0)\\\\h = 1.928 \ m

The work done by gravity on the boy is calculated as;

W = P.E = mgh

             = 32kg x 9.8m/s² x 1.928m

             = 604.62 J

Therefore, the work done by gravity on the boy is 604.62 J

8 0
3 years ago
Two boxes on opposite ends of a massless board that is 3.0 m long. The board is supported in the middle by a fulcrum. The box on
rosijanka [135]

Answer:

b. 1.1 m

Explanation:

It is given that the total distance between the masses is equal to the length of the board, which is 3 m. Therefore,

s_{1} + s_{2} = 3\ m\\\\s_{2} = 3\ m - s_{1}\ --------- eqn(1)

where,

s₁ = distance of fulcrum from left mass

s₂ = distance of fulcrum from right mass

In order to achieve balance, the torque due to both masses must be equal:

T_{1} = T_{2}\\m_{1}s_{1} = m_{2}s_{2}\\(25\ kg)(s_{1}) = (15\ kg)(s_{2})\\\\\frac{15\ kg}{25\ kg}(s_{2}) = s_{1}\\\\using\ eqn(1):\\(0.6)(3\ m - s_{1}) = s_{1}\\1.8\ m = 1.6\ s_{1}\\s_{1} = \frac{1.8\ m}{1.6}

s₁ = 1.1 m

Hence, the correct option is:

<u>b. 1.1 m</u>

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