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3 years ago

Kinetic energy of 110 j and velocity of 70 m/s what is the mass

Physics

1 answer:

VARVARA [1.3K]3 years ago

7 0

Answer:

0.04kg

Explanation:

Explanation in pics:

The formula for kinetic energy is KE = 1/2mv². So now the question asks for the mass, not KE. So we need to rearrange the formula so that we are calculating the mass. So, the formula for mass is m=KE/(1/2v²). Now we just substitute the values into the formula. Then just follow what it shows in the pic

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A conical container of radius 6 ft and height 24 ft is filled to a height of 19 ft of a liquid weighing 64.4 lb divided by ft cu

kobusy [5.1K]

Answer:

Part (i) work required to pump the contents to the rim is 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5ft above the cone's rim is 426,484.878 lb.ft

Explanation:

The center of mass of a uniform solid right circular cone of height h lies on the axis of symmetry at a distance of h/4 from the base and 3h/4 from the top.

Center mass of the liquid Z = (24-19)ft + 19/4 = 5ft + 4.75ft = 9.75 ft

Mass of liquid in the cone = volume × density (ρ) = ¹/₃.π.r².h.ρ

where;

r is the radius of the liquid surface = [6*(19/24)]ft = 4.75ft

ρ is the density of liquid = 64.4 lb/ft³

h is the height of the liquid = 19 ft

Mass of liquid in the cone = ¹/₃ × π × (4.75)² × 19 × 64.4 = 28,914.229 lbs

Part (i) work required to pump the contents to the rim

Work required = 28,914.229 lbs × 9.75 ft = 281,913.733 lb.ft

Part (ii) work required to pump the liquid to a level of 5 ft above the cone's rim

Extra work required = 28,914.229 lb × 5ft = 144571.145 lb.ft

Total work required = (281,913.733 + 144571.145) lb.ft

= 426,484.878 lb.ft

5 0

3 years ago

HURRY ! THE BEST I WILL GIVE THE BRAINLIEST

erik [133]

The speed of both cars is the same ... 80 km per hour.

But their velocities are different, because DIRECTION is part of velocity, and their directions are different.

6 0

3 years ago

Derive equation of motion s=ut+1/2at²

Pavel [41]

Recall the definitions of

• average velocity:

v[ave] = ∆x/∆t = (x[final] - x[initial])/t

Take the initial position to be the origin, so x[initial] = 0, and we simply write x[final] = s. So

v[ave] = s/t

• average acceleration:

a[ave] = ∆v/∆t = (v[final] - v[initial])/t

Assume acceleration is constant (a[ave] = a). Let v[initial] = u and v[final] = v, so that

a = (v - u)/t

Under constant acceleration, the average velocity is also given by

v[ave] = (v[final] + v[initial])/2 = (v + u)/2

Then

v[ave] = s/t = (v + u)/2 ⇒ s = (v + u) t/2

and

a = (v - u)/t ⇒ v = u + at

so that

s = ((u + at) + u) t/2

s = (2u + at) t/2

s = ut + 1/2 at²

4 0

2 years ago

Two point charges of +20.0 μC and -8.00 μC are separated by a distance of 20.0 cm. What is the intensity of electric field E mid

algol13

Answer:

The intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

Explanation:

Here we need first find the electric field due to the first charge at the midway point.

The electric field equation is given by:

$|E_{1}|=k\frac{q_{1}}{d^{2}}$

Where:

k is Coulomb's constant
q(1) is 20.00 μC or 20*10⁻⁶ C
d is the distance from q1 to the midpoint (d=10.0 cm)

So, we will have:

$|E_{1}|=(9*10^{9})\frac{20*10^{-6}}{0.1^{2}}$

$|E_{1}|=1.8*10^{7}\: N/C$

The direction of E1 is to the right of the midpoint.

Now, the second electric field is:

$|E_{2}|=k\frac{q_{2}}{d^{2}}$

$|E_{2}|=(9*10^{9})\frac{8*10^{-6}}{0.1^{2}}$

$|E_{2}|=7.2*10^{6}\: N/C$

The direction of E2 is to the right of the midpoint because the second charge is negative.

Finally, the intensity of the net electric field will:

$E_{net}=E_{1}+E_{2}=2.52*10^{7}\: N/C$

I hope it helps you!

5 0

3 years ago

A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =

Mila [183]

Answer:

$v=6.65m/sec$

Explanation:

From the Question we are told that:

Mass $m=97.6$

Coefficient of kinetic friction $\mu k=0.555$

Generally the equation for Frictional force is mathematically given by

$F=\mu mg$

$F=0.555*97.6*9.8$

$F=531.388N$

Generally the Newton's equation for Acceleration due to Friction force is mathematically given by

$a_f=-\mu g$

$a_f=-0.555 *9.81$

$a_f=-54455m/sec^2$

Therefore

$v=u-at$

$v=0+5.45*1.22$

$v=6.65m/sec$

4 0

3 years ago