In 6 days, the temperature dropped 18 degrees. How many degrees per day did the temperature drop?

Two cars, initially at rest and 5 km apart at t=0 , simultaneously move toward each other. Car A travels at a constant speed of

Anastasy [175]

Answer:

d. 268 s

Explanation:

Constant Speed Motion

An object is said to travel at constant speed if the ratio of the distance traveled by the time taken is constant.

Expressed in a simple equation, we have:

$\displaystyle v=\frac{d}{t}$

Where

v = Speed of the object

d = Distance traveled

t = Time taken to travel d.

From the equation above, we can solve for d:

d = v . t

And we can also solve it for t:

$\displaystyle t=\frac{d}{v}$

Two cars are initially separated by 5 km are approaching each other at relative speeds of 55 km/h and 12 km/h respectively. The total speed at which they are approaching is 55+12 = 67 km/h.

The time it will take for them to meet is:

$\displaystyle t=\frac{5}{67}$

t = 0.0746 hours

Converting to seconds: 0.0746*3600 = 268.56

The closest answer is d. 268 s

8 0

3 years ago

The system below has a friction force of 25 N acting on the cart which 8 kg. The mass hanging off the edge has a mass of 6 kg. F

photoshop1234 [79]

The cart will be pulled to the right by the hanging mass, so by Newton's second law, the net force on the cart is

T - 25 N = (8 kg) a

where T is the tension in the rope and a is the acceleration.

The hanging mass has a net force of

(6 kg) g - T = (6 kg) a

where g = 9.8 m/s².

Adding these equations together eliminates T, and we can solve for a :

(T - 25 N) + ((6 kg) g - T ) = (14 kg) a

33.8 N = (14 kg) a

a = (33.8 N) / (14 kg) ≈ 2.4 m/s²

Then the tension in the rope is

T - 25 N = (8 kg) (2.4 m/s²)

T ≈ 25 N + 19.31 N ≈ 44 N

5 0

3 years ago

A monatomic gas is adiabatically compressed to 0.250 of its initial volume. Do each of the following quantities change?

Len [333]

Answer:

Given that

V2/V1= 0.25

And we know that in adiabatic process

TV^န-1= constant

So

T1/T2=( V1 /V2)^ န-1

So = ( 1/0.25)^ 0.66= 2.5

Also PV^န= constant

So P1/P2= (V2/V1)^န

= (1/0.25)^1.66 = 9.98

A. RMS speed is

Vrms= √ 3RT/M

But this is also

Vrms 2/Vrms1= (√T2/T1)

Vrms2=√2.5= 1.6vrms1

B.

Lambda=V/4π√2πr²N

So

Lambda 2/lambda 1= V2/V1 = 0.25

So the mean free path can be inferred to be 0.25 times the first mean free path

C. Using

Eth= 3/2KT

So Eth2/Eth1= T2/T1

So

Eth2= 2.5Eth1

D.

Using CV= 3/2R

Cvf= Cvi

So molar specific heat constant does not change

5 0

3 years ago

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What minimum speed must the rocket have just before impact in order to save the explorer's life?

34kurt

Answer: Assuming that I understand the geometry correctly, the combine package-rocket will move off the cliff with only a horizontal velocity component. The package will then fall under gravity traversing the height of the cliff (h) in a time T given by h = 0.5*g*T^2 However, the speed of the package-rocket system must be sufficient to cross the river in that time v2 = L/T Conservation of momentum says that m1*v1 = (m1 + m2)*v2 where m1 is the mass of the rocket, v1 is the speed of the rocket, m2 is the mass of the package, and v2 is the speed of the package-rocket system. Expressing v2 in terms of v1 v2 = m1*v1/(m1 + m2) and then expressing the time in terms of v1 T = (m1 + m2)*L/(m1*v1) substituting T in the first expression h = 0.5*g*(m1 + m2)^2*L^2/(m1*v1)^2 solving for v1, the speed before impact is given by v1 = sqrt(0.5*g/h)*(m1 + m2)*L/m1

8 0

3 years ago

A. If you set a cannonball to 1500 m/s, what happens? Explain why you think it moves this way.

creativ13 [48]

Answer:

I beleive it would shoot very far up into the sky

Explanation:

7 0

4 years ago

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