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3 years ago

13

A car accelerates from rest at a constant rate 5m/s^2. Which of the following statements is true

1 answer:

dusya [7]3 years ago

8 0

Answer:

jjhbkvjuhigjgyihbgtimnvyuoibc

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How can stretching affect the range of motion of the neck? Hypothesis

kari74 [83]

Answer:

reduce passive stiffness and increase range of movement during exercise.

Explanation:

stretching performed as part of a warm up prior to exercise is thought to reduce passive stiffness and increase range of movement during exercise. in general it appears that is static stretching is most beneficial for athletes requiring flexibility for their sports.

7 0

2 years ago

At constant volume, the heat of combustion of a particular compound is − 3550.0 kJ / mol. When 1.075 g of this compound ( molar

swat32

Answer:

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

Explanation:

First of all let's define the specific molar heat capacity.

$C = \frac{-Q}{n\cdot \Delta T}$ (1)

Where:

Q is the released heat by the system

n is the number of moles

ΔT is the difference of temperature of the system

Now, we can find n with the molar mass (M) the mass of the compound (m).

$n=\frac{m}{M}=6.95\cdot 10^{-3} moles$

Using (1) we have:

$C=\frac{-3550}{6.95\cdot 10^{-3} 4.073}$

$C=1,25\cdot 10^{5} kJ/^{\circ}C$

I hope it helps!

6 0

3 years ago

A bar magnet was placed underneath a sheet of paper where a pile of iron filings sits. In the presence of the energy stored in t

Naya [18.7K]

Answer: 1. The field energy will increase

2. The energy increases, and the lines of force are denser

3. It points toward the field of earths magnetic poles

4. 1 and 2 only

5. 2, 4, 1, 3

Explanation: just took it

4 0

2 years ago

The sun rotates once in 25.05 days. That is 601.2 hours. so the sun's rotational speed is 0.0017 rotations per hour. what is the

Margaret [11]

Answer:

$v=2019.09\ m.s^{-1}$

Explanation:

Given:

time taken by the sun to complete one revolution, $t=601.2\ hr$

radial distance of the sunspot, $r=6.955\times 10^8\ m.s^{-1}$

<u>Therefore, angular speed of rotation of sun:</u>

$\omega=\frac{2\pi}{601.2\times 3600} \ rad.s^{-1}$

<u>Now the tangential velocity of the sunspot can be given by:</u>

$v=r.\omega$

$v=6.955\times 10^8\times \frac{2\pi}{601.2\times 3600}$

$v=2019.09\ m.s^{-1}$

4 0

3 years ago

Please help me question 8

Rudik [331]

I beleive that the answer is B.

7 0

3 years ago