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3 years ago

A car accelerate uniformly from rest at 5m/s2 . Determine it's speed after 10s

Physics

2 answers:

krek1111 [17]3 years ago

6 0

Answer:

50m/s.

Explanation:

Let's take acceleration as A and speed as S:

A = 5m/s²

S = A × 10s = 5 × 10 = 50m/s

The answer is 50m/s.

Sergeu [11.5K]3 years ago

4 0

Answer:

50m/s

Explanation:

Given:

a=5m/s^2

t=10s

Required:

v=?

Formula:

a=v/t

Solution:

a=v/t

v=a*t

v=5m/s^2*10s

v=50m/s

Hope this helps ;)❤❤❤

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Shalnov [3]

Answer:

Explanation:

During the first .8 s , the elevator is under acceleration . It starts from initial velocity u = 0 , final velocity v = 1.2 m /s , time = .8 s

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3 years ago

Calculate the orbital period of a dwarf planet found to have a semimajor axis of a = 4.0x 10^12 meters in seconds and years.

padilas [110]

Explanation:

We have,

Semimajor axis is $4\times 10^{12}\ m$

It is required to find the orbital period of a dwarf planet. Let T is time period. The relation between the time period and the semi major axis is given by Kepler's third law. Its mathematical form is given by :

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G is universal gravitational constant

M is solar mass

Plugging all the values,

$T^2=\dfrac{4\pi ^2}{6.67\times 10^{-11}\times 1.98\times 10^{30}}\times (4\times 10^{12})^3\\\\T=\sqrt{\dfrac{4\pi^{2}}{6.67\times10^{-11}\times1.98\times10^{30}}\times(4\times10^{12})^{3}}\\\\T=4.37\times 10^9\ s$

Since,

$1\ s=3.17\times 10^{-8}\ \text{years}\\\\4.37\times 10^9\ s=4.37\cdot10^{9}\cdot3.17\cdot10^{-8}\\\\4.37\times 10^9\ s=138.52\ \text{years}$

So, the orbital period of a dwarf planet is 138.52 years.

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3 years ago

a flag of mass 2.5 kg is supported by a single rope. A strong horizontal wind exerts a force of 12 N on the flag. Calculate the

tatuchka [14]

The free-body diagram of the forces acting on the flag is in the picture in attachment.

We have: the weight, downward, with magnitude
$W=mg = (2.5 kg)(9.81 m/s^2)=24.5 N$
the force of the wind F, acting horizontally, with intensity
$F=12 N$
and the tension T of the rope. To write the conditions of equilibrium, we must decompose T on both x- and y-axis (x-axis is taken horizontally whil y-axis is taken vertically):
$T \cos \alpha -F=0$
$T \sin \alpha -W=$
By dividing the second equation by the first one, we get
$\tan \alpha = \frac{W}{F}= \frac{24.5 N}{12 N}=2.04$
From which we find
$\alpha = 63.8 ^{\circ}$
which is the angle of the rope with respect to the horizontal.

By replacing this value into the first equation, we can also find the tension of the rope:
$T= \frac{F}{\cos \alpha}= \frac{12 N}{\cos 63.8^{\circ}}=27.2 N$

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3 years ago

An unknown material, m1 = 0.49 kg, at a temperature of T1 = 92 degrees C is added to a Dewer (an insulated container) which cont

erastova [34]

Answer:

$c_u=1540.5J/kg^{\circ}K$

Explanation:

We know that heat relates to mass, specific heat and variation of temperature experimented because of this heat through the equation $Q=mc\Delta T=mc(T_f-T_i)$ . The heat released by the unknown material is absorbed by water, so we have $Q_u=-Q_w$ , and we can write:

$m_uc_u(T_{uf}-T_{ui})=-m_wc_w(T_{wf}-T_{wi})$

Since thermal equilibrium is reached we know that $T_{cf}=T_{wf}=T_f=31^{\circ}C=304^{\circ}K$ , where we have added $273^{\circ}$ to convert the temperature from Celsius to Kelvin, as we must do. Since we want the specific heat of the unknown material, we do:

$c_u=-\frac{m_wc_w(T_f-T_{wi})}{m_u(T_f-T_{ui})}$

Which for our values is:

$c_u=-\frac{(1.1kg)(4186J/kg^{\circ}K)((304^{\circ}K)-(294^{\circ}K))}{(0.49kg)((304^{\circ}K)-(365^{\circ}K))}=1540.5J/kg^{\circ}K$

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3 years ago

A car accelerate uniformly from rest at 5m/s2 . Determine it's speed after 10s​

A car accelerate uniformly from rest at 5m/s2 . Determine it's speed after 10s