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Monica [59]
4 years ago
15

A runner has an initial velocity of 4 meters per second. After 20 seconds, the runner's velocity is 6 meters per second. Which i

s the runner's acceleration?
Physics
2 answers:
Vikentia [17]4 years ago
7 0
<span>6-4)mps/(20sec) = .1 m/sec^2<span>
</span></span>
Anuta_ua [19.1K]4 years ago
4 0
0.1 m/s^2 (final velocity-initial velocity) and then divide with the time taken
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Two protons are maintained at a separation of nm. Calculate the electric potential due to the two particles at the midpoint betw
Liono4ka [1.6K]

Answer:

The electric potential is approximately 5.8 V

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero

Explanation:

The two protons can be considered as point charges. Therefore, the electric potential is given by the point charge potential:

\displaystyle{U=\frac{q}{4\pi \epsilon_0r}} (1)

where q is the charge of the particle, \epsilon_0 the electric permittivity of the vacuum (I assuming the two protons are in a vacuum) and r is the distance from the point charge to the point where the potential is being measured. Because the electric potential is an scalar, we can simply add the contribution of the two potentials in the midpoint between the protons. Thus:

\displaystyle{U_{midpoint}=\frac{q}{4\pi \epsilon_0r}}+\frac{q}{4\pi \epsilon_0r}}=\frac{q}{2\pi \epsilon_0r}}}

Substituting the values q=1.602 \cdot10^{-19}\ C, \displaystyle{\frac{1}{4\pi\epsilon_0}=8.99\cdot 10^9 N\cdot m^2\cdot C^{-2}} and r=0.5 \cdot 10^{-9} m we obtain:

\displaystyle{U_{midpoint}=\frac{q}{2\pi \epsilon_0r}}=5.759 \approx 5.8 V}

The resulting direction of the electric field will lie on the line that joins the charges but since it is calculated in the midpoint and the charges are the same we can directly say that its magnitude is zero.

6 0
4 years ago
The average specific heat of the human body is 3.6 kJ/kg·°C. If the body temperature of a(n) 96-kg man rises from 37°C to 39°C d
d1i1m1o1n [39]

Answer:

691200 J

Explanation:

From specific heat capacity,

ΔQ = cmΔt.................. Equation 1

Where ΔQ = increase in thermal energy, c = specific heat capacity of the body, m = mass of the man, Δt = rise in temperature.

Given: c = 3.6 kJ/kg.°C = 3600 J/kg.°C, m = 96 kg, Δt = 39-37 = 2 °C.

Substitute into equation 1

ΔQ = 3600×96×2

ΔQ = 691200 J.

Hence the change in the thermal energy of the body = 691200 J

5 0
3 years ago
A(n) 1400-kg car going at 6.32 m/s in the positive x direction collides with a 2900-kg truck at rest. The collision is totally i
tresset_1 [31]

Answer:

a) Acceleration of the car is given as

a_{car} = -21 m/s^2

b) Acceleration of the truck is given as

a_{truck} = 10.15 m/s^2

Explanation:

As we know that there is no external force in the direction of motion of truck and car

So here we can say that the momentum of the system before and after collision must be conserved

So here we will have

m_1v_1 + m_2v_2 = (m_1 + m_2)v

now we have

1400 (6.32) + 2900(0) = (1400 + 2900) v

v = 2.06 m/s

a) For acceleration of car we know that it is rate of change in velocity of car

so we have

a_{car} = \frac{v_f - v_i}{t}

a_{car} = \frac{2.06 - 6.32}{0.203}

a_{car} = -21 m/s^2

b) For acceleration of truck we will find the rate of change in velocity of the truck

so we have

a_{truck} = \frac{v_f - v_i}{t}

a_{truck} = \frac{2.06 - 0}{0.203}

a_{truck} = 10.15 m/s^2

5 0
3 years ago
What is the ability of the body to function successfully and efficiently during
jolli1 [7]
I believe this question should be in Health, however, I'll answer it to the best of my ability.

Answer: Physical fitness.  

Physical fitness can be achieved through a set diet, physical excersise (of course), rest, and many cardiovascular exercises. It is the ability of the body to successfully function efficiently during PA. 
8 0
4 years ago
A spring-loaded toy dart gun is shot to a height h. The same dart is shot straight up a second time from the same gun, but this
yarga [219]

Answer:

The height reached by the dart in the second shot is (4 H).

Explanation:

It is given that, a spring-loaded toy dart gun is shot to a height h. In this case, all the potential energy stored in the spring is converted to potential gravitational energy at the maximum height.

\dfrac{1}{2}kx^2=mgH........(1)

At the second shot, the spring is compressed twice as far before firing. x' = 2x

\dfrac{1}{2}kx'^2=mgh

\dfrac{1}{2}k(2x)^2=mgh.........(2)

h is the height reached by the dart in the second shot.

Dividing equation (1) and (2) as:

4=\dfrac{h}{H}

h = 4H

So, the height reached by the dart in the second shot is (4 H). Hence, this is the required solution.

8 0
4 years ago
Read 2 more answers
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