What would be the final temperature if you mixed a liter of 40C water with 2 liters of 20C water?

A 15g bullet is fired horizontally into a 3kg block of wood suspended by a long cord. Assume that the bullet remains in the bloc

Varvara68 [4.7K]

Answer:

261.3 m/s

Explanation:

Mass of bullet=m=15 g= $\frac{15}{1000}=0.015 kg$

1 kg=1000g

Mass of block=M=3 kg

d=0.086 m

Total mass =M+m=3+0.015=3.015 kg

K.E at the time strike=Gravitational potential energy at the end of swing

$\frac{1}{2}(m+M)^2V^2=(m+M)gh$

Using g= $9.8m/s^2$

Substitute the values

$\frac{1}{2}(3.015)V^2=3.015\times 9.8\times 0.086$

$V^2=\frac{2\times 3.015\times 9.8\times 0.086}{3.015}$

$V=\sqrt{2\times 3.015\times 9.8\times 0.086}}$

$V=1.3m/s$

Velocity after collision=V=1.3 m/s

Velocity of block=v'=0

Using conservation law of momentum

$mv+Mv'=(m+M)V$

Using the formula

$0.015v+3(0)=3.015(1.3)$

$0.015v=3.015(1.3)$

$v^2=\frac{3.015(1.3)}{0.015}=261.3$

$v=261.3 m/s$

5 0

3 years ago

A brass alloy is known to have a yield strength of 240 MPa (35,000 psi), a tensile strength of 310 MPa (45,000 psi), and an elas

Karo-lina-s [1.5K]

Answer:

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

Explanation:

Given that

Yield strength ,Sy= 240 MPa

Tensile strength = 310 MPa

Elastic modulus ,E= 110 GPa

L=380 mm

ΔL = 1.9 mm

Lets find strain:

Case 1 :

Strain due to elongation (testing)

ε = ΔL/L

ε = 1.9/380

ε = 0.005

Case 2 :

Strain due to yielding

$\varepsilon' =\dfrac{S_y}{E}$

$\varepsilon' =\dfrac{240}{110\times 1000}$

ε '=0.0021

Here Strain due to testing is greater than the strain due to yielding that is why computation of load is not possible.

For computation of load strain due to testing should be less than the strain due to yielding.

4 0

3 years ago

What is the mass of a truck in grams of it produces a force of 1500N while accelerating at a rate of 6 m/s²?

aleksley [76]

Answer:

250,000

Explanation:

<h2>formula = ( F=ma </h2>

F=1500N
a=6m/s^2
F= ma
m=?
1500/6 = m
m=250 kg
1kg =1000gm so 250kg =250,000gm
m =250×10^3 gm

5 0

3 years ago

A rock is being twirled in a circle on the end of a string. The string provides the centripetal force needed to keep the ball mo

KonstantinChe [14]

Answer:

No

Explanation:

The force of tension exerted by the string on the rock acts as centripetal force, so its direction is always towards the centre of the circle.

However, the direction of motion of the rock is always tangential to the circle: this means that the force is always perpendicular to the direction of motion of the rock.

As we know, the work done by a force on an object is

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the force and the displacement

In this situation, F and d are perpendicular, so $\theta=90^{\circ}$ , therefore $cos \theta = 0$ and the work done is zero:

$W=0$

4 0

2 years ago

A van is traveling on a road at a speed of 55 km/h. A girl sitting near the driver of the van throws a paper airplane to a boy a

aivan3 [116]

<span>A van is traveling on a road at a speed of 55 km/h relative to a
stationary observer on the side of the road. A girl sitting near the
driver of the van throws a paper airplane to a boy at the back of the
van with a speed of 2 km/h relative to the girl, the boy, and the van.

The speed of the paper airplane, relative to the same stationary observer
on the side of the road, is (55 - 2) = 53 km/h. No rounding is necessary.</span>

5 0

3 years ago

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