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3 years ago

Do you think most people follow the recommended guidelines for an adequate exercise routine?

1 answer:

6 0

Personally, I know that I don't however it all depends on the person's lifestyle, health, and resoucres, such as if they are able to complete the excersise or have enough money fore equpemnt. To be safe, I would say that the majority of people do follow the guidelines although they may manipulate it to fit their best interest.

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Pete Zaria works on weekends at Barnaby's Pizza Parlor. His primary responsibility is to fill drink orders for customers. He fil

laila [671]

Answer:

$W_n_e_t=7.648512 \approx 7.6J$

$K.E=0.8J$

$v=0.7844645406 \approx 0.78m/s$

Explanation:

From the question we are told that

Mass of pitcher $M= 2.6kg$

Force on pitcher $f=8.8N$

Distance traveled $48cm=>0.48m$

Coefficient of friction $\mu=0.28$

a)Generally frictional force is mathematically given by

$F=\mu N$

$F=0.28*2.6*9.8$

$F=7.1344N$

Generally work done on the pitcher is mathematically given as

$W_n_e_t=W_f+W_F$

$W_f=8.8*0.48=>4.224N\\W_F=7.1344*0.48=>3.424512N$

$W_n_e_t=4.224-3.424512$

$W_n_e_t=0.799488\approx 0.8J$

b)Generally K.E can be given mathematically as

$K.E= W_n_e_t$

Therefore

$K.E=0.8J$

c)Generally the equation for kinetic energy is mathematically represented by

$K.E=1/2mv^2$

$0.8=1/2mv^2$

Velocity as subject

$v=\sqrt{\frac{K.E*2}{m} }$

$v=\sqrt{\frac{0.8*2}{2.6} }$

$v=0.7844645406 \approx 0.78m/s$

6 0

3 years ago

Over a period of operation, the useful work output of the fluorescent bulb was

Nadya [2.5K]

Answer:

199.0521 Will be the answer

5 0

3 years ago

Write SHORT examples of Gravitational potential energy and Elastic energy

Nikolay [14]

Dropping a bouncy ball and stretching a rubber ban.

6 0

3 years ago

Read 2 more answers

if 100 kilojoules of energy is used to heat 500 grams of water what is the temperature change of the water

Serjik [45]

Answer:

47.8 °C

Explanation:

Use the heat equation:

q = mCΔT

where q is the heat absorbed/lost,

m is the mass of water,

C is the specific heat capacity,

and ΔT is the change in temperature.

Here, q = 100 kJ, m = 0.5 kg, and C = 4.184 kJ/kg/°C.

100 kJ = (0.5 kg) (4.184 kJ/kg/°C) ΔT

ΔT = 47.8 °C

6 0

3 years ago

A wheel rotating about a fixed axis has a constant angular acceleration of 4.0 rad/s2. In a 4.0-s interval the wheel turns throu

anastassius [24]

Answer:

a. 3 s.

Explanation:

Given;

angular acceleration of the wheel, α = 4 rad/s²

time of wheel rotation, t = 4 s

angle of rotation, θ = 80 radians

Apply the kinematic equation below,

$\theta = \omega_1 t \ + \ \frac{1}{2} \alpha t^2\\\\80 = 4\omega_1 + \frac{1}{2}*4*4^2\\\\80 = 4\omega_1 + 32\\\\ 4\omega_1 = 48\\\\ \omega_1 = \frac{48}{4}\\\\ \omega_1 = 12 \ rad/s$

Given initial angular velocity, ω₀ = 0

Apply the kinematic equation below;

$\omega_1 = \omega_o + \alpha t_1\\\\12 = 0 + 4t\\\\4t = 12\\\\t = \frac{12}{4}\\\\t = 3 \ s$

Therefore, the wheel had been in motion for 3 seconds.

a. 3 s.

8 0

3 years ago