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Zina [86]
3 years ago
9

A substance freezes at -58c therefore substance melts at ?

Physics
1 answer:
Yanka [14]3 years ago
3 0
The freezing point is the same as the melting point.

If it freezes at -58°C, hence the melting point is also <span>-58°C.</span>
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Calculate the kinetic energy in joules of a 1,500 kg car that is moving at a speed of 42 km/h
Rzqust [24]
Data:
KE (Kinetic Energy) = ? (Joule)
m (mass) = 1500 Kg 
v (speed) = 42 Km/h
converting to m/s (42 / 3.6), we have: v (speed) = 11.6 m/s

Formula:
K_{E} =   \frac{1}{2} m*v^2

Solving:
K_{E} = \frac{1}{2} m*v^2
K_{E} =  \frac{1}{2} *1500*(11.6)^2
K_{E} = \frac{1}{2} *1500*134.56
K_{E} =  \frac{201840}{2}
\boxed{\boxed{K_{E} = 100920\:Joule}}\end{array}}\qquad\quad\checkmark





4 0
3 years ago
What would happen if the moon was hit by an asteroid
Semmy [17]
If the moon was hit by an asteroid there would be a crater mark and possible movement.
8 0
4 years ago
A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(
Natasha2012 [34]

Answer:

a) t=2s

b) h_{max}=1024ft

c) v_{y}=-256ft/s

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2

To find what time does it take to reach maximum height we need to find how high will it go

b) We can calculate its initial height using the following formula

Knowing that its velocity is zero at its maximum height

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

0=(64ft/s)^2-2(32ft/s^2)(y-960ft)

y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

h=-16t^2+64t+960

1024=-16t^2+64t+960

0=-16t^2+64t-64

Solving the quadratic equation

t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}

a=-16\\b=64\\c=-64

t=2s

So, the projectile reach maximum point at t=2s

c) We can calculate the final velocity by using the following formula:

v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})

v_{y}=±\sqrt{(64ft/s)^{2}-2(32ft/s^2)(-960ft)}=±256ft/s

Since the projectile is going down the velocity at the instant it reaches the ground is:

v=-256ft/s

5 0
3 years ago
Type the correct answer in the box. Express your answer to three significant figures.
Marrrta [24]

The mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

<h3>What is mass?</h3>

Mass is a numerical measure of inertia, which is a basic feature of all matter. It is, in effect, a body of matter's resistance to a change in speed or position caused by the application of a force.

In the International System of Units (SI), the kilogram is the unit of mass.

Given;

Mass of calcium nitrate (Ca(NO₃)₂) = 96.1 g

Mass of calcium phosphate=?

The chemical equation is found as;

3Ca(NO₃)₂ + 2Na₃PO₄ → 6NaNO₃ + Ca₃(PO4)₂

3 moles of Ca(NO₃)₂ produces 1 mole of Ca₃(PO4)molar mass of calcium phosphate is 164 g/mol

Moles are found as the ratio of the mass of the compound and its molar mass. Moles of Ca(NO₃)₂ is n.

n = 96.1/164 = 0.5859 moles

Moles of Ca3(PO4)2 produced ;

N=0.0589 ×(1/3) = 0.0196 moles

The molar mass of calcium phosphate is 310 g/mol and the mass of calcium phosphate produced will be;

M=0.0196×310

M = 6.076 g

Hence,the mass of calcium phosphate the reaction can produce is <u>6.076</u> grams.

To learn more about the mass, refer to the link;

brainly.com/question/13073862

#SPJ1

5 0
2 years ago
A train travels due north in a straight line with a constant speed of 100 m/s. Another train leaves a station 2,881 m away trave
damaskus [11]

Answer:

The trains will collide at a distance 1660 m from the station

Explanation:

Let the train traveling due north with a constant speed of 100 m/s be Train A.

Let the train traveling due south with a constant speed of 136 m/s be Train B.

From the question, Train B leaves a station 2,881 m away (that is 2,881 m away from Train A position).

Hence, the two trains would have traveled a total distance of 2,881 m by the time they collide.

∴ If train A has covered a distance x m by the time of collision, then train B would have traveled (2881 - x) m.

Also,

At the position where the trains will collide, the two trains must have traveled for equal time, t.

That is, At the point of collision,

t_{A} = t_{B}

t_{A} is the time spent by train A

t_{B} is the time spent by train B

From,

Velocity = \frac{Distance }{Time }\\

Time = \frac{Distance}{Velocity}

Since the time spent by the two trains is equal,

Then,

\frac{Distance_{A} }{Velocity_{A} }  = \frac{Distance_{B} }{Velocity_{B} }

{Distance_{A} = x m

{Distance_{B} = 2881 - x m

{Velocity_{A} = 100 m/s

{Velocity_{B} = 136 m/s

Hence,

\frac{x}{100} = \frac{2881 - x}{136}

136(x) = 100(2881 - x)\\136x = 288100 - 100x\\136x + 100x = 288100\\236x = 288100\\x = \frac{288100}{236} \\x = 1220.76m\\

x≅ 1,221 m

This is the distance covered by train A by the time of collision.

Hence, Train B would have covered (2881 - 1221)m = 1660 m

Train B would have covered 1660 m by the time of collision

Since it is train B that leaves a station,

∴ The trains will collide at a distance 1660 m from the station.

7 0
3 years ago
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