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belka [17]
3 years ago
9

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

Physics
1 answer:
blondinia [14]3 years ago
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
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