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2 years ago

The speed limit sign in indiana reads 60 miles per hour. convert these units to meters per second

Physics

1 answer:

cluponka [151]2 years ago

3 0

Answer:

26.822 m/s

Explanation:

60 mi/hr * 5280 ft/mile * 1 hr / 3600 sec * 12 in / foot * 1 meter / 39.37 in = <u>26.822 m/s</u>

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Calculate the amount of heat needed to raise the temperature of 200g of ice from-30°C 50C water. (C = .5 for ice, C 1 for water,

Nitella [24]

<u>Answer:</u> The amount of heat needed is 29000 Cal.

<u>Explanation:</u>

The process involved in this problem are:

$(1):H_2O(s)(-30^oC)\rightarrow H_2O(s)(0^oC)\\\\(2):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(3):H_2O(l)(0^oC)\rightarrow H_2O(l)(50^oC)$

Now, we calculate the amount of heat released or absorbed in all the processes.

<u>For process 1:</u>

$q_1=mC_{p,s}\times (T_2-T_1)$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of ice = 200 g

$T_2$ = final temperature = $0^oC$

$T_1$ = initial temperature = $-30^oC$

Putting all the values in above equation, we get:

$q_1=200g\times 0.5Cal/g^oC\times (0-(-30))^oC=3000Cal$

<u>For process 2:</u>

$q_2=m\times L_f$

where,

$q_1$ = amount of heat absorbed = ?

m = mass of water or ice = 200 g

$L_f$ = latent heat of fusion = 80 Cal/g

Putting all the values in above equation, we get:

$q_2=200g\times 80Cal/g=16000Cal$

<u>For process 3:</u>

$q_3=m\times C_{p,l}\times (T_{2}-T_{1})$

where,

$q_3$ = amount of heat absorbed = ?

m = mass of water = 200 g

$T_2$ = final temperature = $50^oC$

$T_1$ = initial temperature = $0^oC$

Putting all the values in above equation, we get:

$q_3=200g\times 1Cal/g^oC\times (50-0)^oC=10000Cal$

Calculating the total heat absorbed, we get:

$Q=q_1+q_2+q_3$

$Q=3000+16000+10000=29000Cal$

Hence, the amount of heat needed is 29000 Cal.

7 0

4 years ago

During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau

Thepotemich [5.8K]

Answer:

$T=6.75s$

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on and the second when the rocket's engines is off. So, we need to know the height ( $h_1$ ) that the rocket reaches while its engine is on and we need to know the distance ( $h_2$ ) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

$h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\$

and the final speed is:

$v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T$

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

$v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2$

The sum of these heights will give us the total height, which is known:

$h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s$