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3 years ago

Help!Yall get brainliest

Physics

2 answers:

VLD [36.1K]3 years ago

8 0

Roughage, or fiber, refers to the carbs in plants that your body cannot digest. So meaning false because there is no fiber in meat

Alex_Xolod [135]3 years ago

7 0

Answer:

Explanation:

Good sources include barley, oatmeal, beans, nuts, and fruits such as apples, berries, citrus fruits, and pears. Many foods contain both soluble and insoluble fiber. In general, the more natural and unprocessed the food, the higher it is in fiber. There is no fiber in meat, dairy, or sugar.

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A traffic accident detective measures the skid marks left by a car, 825kg. He determines that the distance between the point tha

deff fn [24]

Doing a force balance on the car:
ma = Fr
ma = μmg
a = μg
a = 0.3(9.81)
a = 29.43 m/s2

Using the formula:
2ax = v2
2(29.43)(34) = v2
v = 44.74 m/s = 161.05 km/h

The car was going 44.74 m/s or 161.05 kph when the brakes were applied.

7 0

3 years ago

A small rock is thrown vertically upward with a speed of 21.0 m/sm/s from the edge of the roof of a 21.0-mm-tall building. The r

FromTheMoon [43]

Answer:

Explanation:

Given:

Vo = 21 m/s

Vf = 0 m/s

Using equation of Motion,

Vf^2 = Vo^2 - 2aS

S = (21^2)/2 × 9.8

= 22.5 m.

Given:

S = 22.5 + 21 mm

= 22.521 m

Vo = 0 m/s

Using the equation of motion,

S = Vo × t + 1/2 × a × t^2

22.521 = 0 + 1/2 × 9.8 × t^2

t^2 = (2 × 22.521)/9.8

= 4.6

t = 2.14 s

4 0

4 years ago

A racing car is travelling at 70 m/s and accelerates at -14 m/s2. What would the car’s speed be after 3 s?

Leno4ka [110]

<h3><u>Question</u><u>:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2. What would the car’s speed be after 3 s?

<h3><u>Statement:</u></h3>

A racing car is travelling at 70 m/s and accelerates at -14 m/s^2.

<h3><u>Solution</u><u>:</u></h3>

Initial velocity (u) = 70 m/s
Acceleration (a) = -14 m/s^2
Time (t) = 3 s
Let the velocity of the car after 3 s be v m/s
By using the formula,

v = u + at, we have

$v = 70 + ( - 14)(3) \\ = > v = 70 - 42 \\ = > v = 28$

So, the velocity of the car after 3 s is 28 m/s.

<h3><u>Answer:</u></h3>

The car's speed after 3 s is 28 m/s.

Hope it helps

3 0

3 years ago

Please help with some/all questions on this worksheet

Roman55 [17]

A. Yes electricity is a fuel this is because protons are being displaced threw currents of electricity that are the used too charge a coil witch spins causing rotationional force And power too appropriate machine
B. Octane
C.one advantage is more power since octane is more combustable than electricity it makes gas or Diesel engines more powerful as too why there are no semi trucks being powered by electricity.
D. One disadvantage is effectively how well its fuel is spent gas is 69.78% less efficient

5 0

2 years ago

A playground merry-go-round has a mass of 120 kg and a radius of 1.80 m and it is rotating with an angular velocity of 0.400 rev

saw5 [17]

Answer:

The final angular velocity is rev/s is 0.293 rev/s.

Explanation:

Given;

mass of the merry-go-round, m₁ = 120 kg

radius of the merry-go-round, r = 1.8 m

initial angular velocity, ω = 0.4 rev/s

mass of the child, m₂ = 22 kg

Apply the principle of conservation angular momentum to determine the final angular velocity;

$I_i= I_f\\\\\frac{1}{2} m_1r^2 \omega _i = \frac{1}{2} m_1r^2 \omega _f + m_2r^2 \omega _f\\\\ \frac{1}{2} m_1r^2 \omega _i =( \frac{1}{2} m_1r^2 + m_2r^2 )\omega _f\\\\\omega _f = \frac{ \frac{1}{2} m_1r^2 \omega _i}{\frac{1}{2} m_1r^2 + m_2r^2} \\\\\omega _f = \frac{ \frac{1}{2} m_1 \omega _i}{\frac{1}{2} m_1 + m_2}\\\\\omega _f = \frac{0.5 \ \times \ 120\ kg \ \times \ 0.4\ rev/s}{0.5 \ \times 120\ kg \ \ + \ \ 22 \ kg} \\\\\omega _f = 0.293 \ rev/s\\$

Therefore, the final angular velocity is rev/s is 0.293 rev/s.

3 0

3 years ago