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4 years ago

A person's temperature is 40°C What would it be in Kelvin?

Physics

1 answer:

frez [133]4 years ago

6 0

Formula :

K = 273.15 + C

40°C

K= 273,15 +40°

K= 313,15 °

40 degrees Celsius = 313.15 kelvin

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A 15 kg cart is pushed on a frictionless surface from rest horizontally by a 30 N force. What is the cart's acceleration?

Rainbow [258]

Answer:

a. The cart's acceleration is 2 m/s^2

b. The cart will travel 100 m

c. The speed is 20 m/s

Explanation:

a. The acceleration of the cart can be calculated using Newton's second law:

F = m.a

Solving for a:

$\displaystyle a=\frac{F}{m}$

The cart has a mass of m=15 Kg and is applied a net force of F=30 N, thus:

$\displaystyle a=\frac{30}{15}$

$a=2\ m/s^2$

Now we use kinematics to find the distance and speed:

$\displaystyle x = v_o.t+\frac{at^2}{2}$

The cart starts from rest (vo=0). The distance traveled in t=10 seconds is:

$\displaystyle x = 0*10+\frac{2*10^2}{2}$

$x = 100\ m$

The cart will travel 100 m

The final speed is calculated by:

$v_f=0+2*10=20\ m/s$

The speed is 20 m/s

5 0

3 years ago

Can someone help me with 6 and 7

natulia [17]

A because the dot nearst to a

5 0

4 years ago

Read 2 more answers

A rock falls from a vertical cliff that is 4.0 m tall and experiences no significant air resistance as it falls. At what speed w

Zinaida [17]

Answer:

v = 6.3 m/s

Explanation:

Since no significant air resistance exists, total mechanical energy must be kept constant at any time.
At the top of the cliff, all the energy is gravitational potential energy, as follows:

$E_{i} = K_{i} + U_{i} = 0 + U_{i} (1)$

If we choose the ground level as our zero reference level for the gravitational potential energy, Ui is simply:
Ui = m*g*h (1)
At any height, the sum of the kinetic and the gravitational potential energy must be equal to (1).
We know from the question, that at the point of interest, both types of energies must be equal each other, so we can write the following expression from (1):

$m*g* h = 2*\frac{1}{2}*m*v^{2} (2)$

Dividing both sides by m, simplifying, and solving for v, we get:

$v = \sqrt{g*h} =\sqrt{9.8m/s2*4.0m} = 6.26 m/s (3)$

v = 6.3 m/s (with two significative figures)

8 0

3 years ago

Find the weight required to stretch a steel road by 2 mm, if the length of the road is 2 m and diameter 4 mm. Y= 200 GPa and g -

svetlana [45]

Answer:

Option D is the correct answer.

Explanation:

We equation for elongation

$\Delta L=\frac{PL}{AE}$

Here we need to find weight required,

We need to stretch a steel road by 2 mm, that is ΔL = 2mm = 0.002 m

$A=\frac{\pi d^2}{4}=\frac{\pi ({4\times 10^{-3}})^2}{4}=1.26\times 10^{-5}m^2$

E = 200GPa = 2 x 10¹¹ N/m²

L=2m

Substituting

$0.002=\frac{P\times 2}{1.26\times 10^{-5}\times 2\times 10^{11}}\\\\P=2520N=256.4kg$

Option D is the correct answer.

6 0

3 years ago

A body goes from P to Q with a velocity of 10 m/s and comes back from Q to P with a velocity of 20 m/s. What is the average velo

Pachacha [2.7K]

Let's say the distance is D. Then the time going is D/10 sec. The time returning is D/20 s. The total time is 3D/20 s, and the total distance is 2D. The average speed for the round trip is (total distance)/(total time). That's (2D) ÷ (3D/20). That's (40D/3D) which is 13-1/3 m/s. (I thought it was going to depend on the distance, but it doesn't.)

5 0

3 years ago