A 40 kg shell is flying at a speed of 72 km/h. It explodes into two

I WILL GIVE BRAINLIEST IF SOMEONE GETS THIS......

pav-90 [236]

Answer:

Explanation:

a)

Firstly to calculate the total mass of the can before the metal was lowered we need to add the mass of the eureka can and the mass of the water in the can. We don't know the mass of the water but we can easily find if we know the volume of the can. In order to calculate the volume we would have to multiply the area of the cross section by the height. So we do the following.

100 $cm^{2}$ x 10cm = 1000 $cm^{3}$

Now in order to find the mass that water has in this case we have to multiply the water's density by the volume, and so we get....

$\frac{1g}{cm^{3} }$ x 1000 $cm^{3}$ = 1000g or 1kg

Knowing this, we now can calculate the total mass of the can before the metal was lowered, by adding the mass of the water to the mass of the can. So we get....

1000g + 100g = 1100g or 1.1kg

b)

The volume of the water that over flowed will be equal to the volume of the metal piece (since when we add the metal piece, the metal piece will force out the same volume of water as itself, to understand this more deeply you can read the about "Archimedes principle"). Knowing this we just have to calculate the volume of the metal piece an that will be the answer. So this time in order to find volume we will have to divide the total mass of the metal piece by its density. So we get....

20g ÷ $\frac{8g}{cm^{3} }$ = 2.5 $cm^{3}$

c)

Now to find out the total mass of the can after the metal piece was lowered we would have to add the mass of the can itself, mass of the water inside the can, and the mass of the metal piece. We know the mass of the can, and the metal piece but we don't know the mass of the water because when we lowered the metal piece some of the water overflowed, and as a result the mass of the water changed. So now we just have to find the mass of the water in the can keeping in mind the fact that 2.5 $cm^{3}$ overflowed. So now we the same process as in number a) just with a few adjustments.

$\frac{1g}{cm^{3} }$ x (1000 $cm^{3}$ - 2.5 $cm^{3}$ ) = 997.5g

So now that we know the mass of the water in the can after we added the metal piece we can add all the three masses together (the mass of the can. the mass of the water, and the mass of the metal piece) and get the answer.

100g + 997.5g + 20g = 1117.5g or 1.1175kg

5 0

3 years ago

Consider a uniform electric field of 50 N/C directed toward the east. If the voltage measured relative to ground at a given poin

qaws [65]

Answer:

30 V

Explanation:

Given that:

The uniform electric field = 50 N/C

Voltage = 80 V

distance = 1.0 m

The potential difference of the electric field = Δ V

E_d = V₁ - V₂

50 × 1 = 80V - V₂

50 - 80 V = - V₂

-30 V = - V₂

V₂ = 30 V

5 0

3 years ago

Cars A and B are racing each other along the same straight road in the following manner: Car A has a head start and is a distanc

4vir4ik [10]

The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance $D_{A}$ beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed $v_{A}$ . Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed $v_{B}$ , which is greater than $v_{A}$ .

Part A: How long after Car B started the race will Car B catch up with Car A? Express the time in terms of given quantities.

Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

Answer: Part A: $t=\frac{D_{A}}{v_{B}-v_{A}}$

Part B: $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

$x=x_{0}+vt$

where

$x_{0}$ is initial position

v is velocity

t is time

Car A started the race at a distance. So at t = 0, initial position is $D_{A}$ .

The equation will be:

$x_{A}=D_{A}+v_{A}t$

Car B started at the starting line. So, its equation is

$x_{B}=v_{B}t$

Part A: When they meet, both car are at "the same position":

$D_{A}+v_{A}t=v_{B}t$

$v_{B}t-v_{A}t=D_{A}$

$t(v_{B}-v_{A})=D_{A}$

$t=\frac{D_{A}}{v_{B}-v_{A}}$

Car B meet with Car A after $t=\frac{D_{A}}{v_{B}-v_{A}}$ units of time.

Part B: With the meeting time, we can determine the position they will be:

$x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )$

$x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$

Since Car B started at the starting line, the distance Car B will be when it passes Car A is $x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}$ units of distance.

5 0

3 years ago

Now we’ll use the component method to add two vectors. We will use this technique extensively when we begin to consider how forc

dimaraw [331]

Answer:

The magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

Explanation:

Since vector A has magnitude 50 cm and a direction of 30, its x - component is A' = 50cos30 = 43.3 cm and its y - component is A" = 50sin30 = 25.

Also, Since vector B has magnitude 35 cm and a direction of 110, its x - component is A' = 35cos110 = -11.97 cm and its y - component is A" = 35sin110 = 32.89 cm.

So, the vector sum R = A + B

The x-component of the vector sum is R' = A'+ B' = 43.3 cm + (-11.97 cm) = 43.3 cm - 11.97 cm = 31.33 cm

The y-component of the vector sum is R" = A"+ B" = 25 cm + 32.89 cm = 57.89 cm

So, the magnitude of R = √(R'² + R"²)

= √((31.33 cm)² + (57.89 cm)²)

= √(981.5689 cm² + 3,351.2521 cm²)

= √(4,332.821 cm²)

= 65.82 cm

≅ 65.8 cm

The direction of R is Ф = tan⁻¹(R"/R')

= tan⁻¹(57.89 cm/31.33 cm)

= tan⁻¹(1.84775)

= 61.58°

≅ 61.6°

So, the magnitude of the vector sum of A and B is 65.8 cm and its direction 61.6°

4 0

3 years ago

A. In one short sentence, explain why we call the force of gravity an attractive force.

kondor19780726 [428]

Answer:

Explanation:

(a) The force of gravity is called an attractive force because it is the force (although weak) in which a planetary body or matter uses to attract an object towards itself.

(b) Yes, it does and the formula for force of gravity between any two object is

F = G $\frac{m1m2}{r}$

where m1 and m2 are masses of the first and second object respectively

r is the distance between the center of the two masses

G is the gravitational constant

3 0

3 years ago