Can a single atom be considered a molecule?

A tow truck drags a stalled car along a road. The chain makes an angle of 30???? with the road and the tension in the chain is 1

My name is Ann [436]

Answer: work = 1,305kJ

Explanation:

angle= 30°

force= 1,500N

distance= 1,000m

The formula for work is : Work= force x distance, however there is an angle of 30° between the direction of force applied and the direction of motion, therefore force must be decomposed to its value on the horizontal axis which is the direction of motion by using the cosine of the very angle.

W= F×cos(α)×D

W= 1,500×cos (30)×1,000

W= 1,305kJ ( kilojoules)

3 0

4 years ago

for the whole page done i reward u with 50 points i am very tired and i have many work to do so thank you guys for the help

mezya [45]

Answer:

Box 1

Explanation:

Formula we are using :

Force = mass × acceleration

or

mass = Force / acceleration (since mass needs to be found)

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Box 1 :

⇒ mass = 5 N / 5 m/s²

⇒ mass = 1 kg

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Box 2 :

⇒ mass = 5 N / 0.75 m/s²

⇒ mass = 5 × 4/3 = 20/3 kg

⇒ mass = 6.67 kg

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Box 3 :

⇒ mass = 5 N / 4.3 m/s²

⇒ mass = 50/43 kg

⇒ mass = 1.16 kg

===========================================================

On comparing Box 1, Box 2, and Box 3, we understand that Box 1 has the smallest mass

4 0

2 years ago

Read 2 more answers

Solve for the missing variable for each set. Showing ALL work through GUESS Method. Given, Unknown, Equation, Setup, Solve. 1. P

Lelechka [254]

Answer:

(1) P = 100 kg-m/s (2) M = 40 kg.

Explanation:

We need to solve the missing variable :

1. P=?, M=20 kg, V=5 m/s

If M is mass, V is velocity, then momentum P is given by :

P = MV

P = 20 kg × 5 m/s

P = 100 kg-m/s

(2) P= 3000 kg(m/s), M=?, V=75 m/s

$M=\dfrac{P}{V}\\\\M=\dfrac{3000}{75}\\\\M = 40\ kg$

Hence, P = 100 kg-m/s and M = 40 kg.

5 0

3 years ago

1. After being struck by a bowling ball, a 1.5 kg bowling pin moving to the right at +3.0 m/s collides with

salantis [7]

Answer:

a) Final velocity of second bowling pin is 2.5m/s.

b) Final velocity of second bowling pin is 3 m/s.

Explanation:

Let 'm' be the mass of both the bowling pin -

m = 1.5 kg

Initial velocity of first bowling pin -

$v_{1} = 3 m/s$

In any type of collision between two bodies in horizontal plane , momentum is conserved along the line of impact.

a) Since , initial velocity of second bowling pin is 0 m/s -

Initial momentum ,

$p_{1} = mv_{1}$

Final velocity of first bowling pin , $v_{2} = 0.5m/s$ [Considering initial direction of motion of the first bowling pin to be positive]

Let $u_{2}$ be the final velocity of the second bowling pin.

∴ Final momentum ,

$p_{2} = mv_{2} + mu_{2}$ .

Now ,

$p_{1} = p_{2}$

∴ $mv_{1} = mv_{2} + mu_{2}$

∴ $u_{2}$ = 3 - 0.5 = 2.5 m/s

∴ Final velocity of second bowling pin is 2.5 m/s.

b) Since , initial velocity of second bowling pin is 0 m/s -

Initial momentum ,

$p_{1} = mv_{1}$

Final velocity of first bowling pin , $v_{2} = 0m/s$ [given][Considering initial direction of motion of the first bowling pin to be positive]

Let $u_{2}$ be the final velocity of the second bowling pin.

∴ Final momentum ,

$p_{2} = mv_{2} + mu_{2}$ .

Now ,

$p_{1} = p_{2}$

∴ $mv_{1} = mv_{2} + mu_{2}$

∴ $u_{2}$ = 3 - 0 = 3 m/s

∴ Final velocity of second bowling pin is 3 m/s.

7 0

3 years ago

Estimate the theoretical fracture strength of a brittle material if it is known that fracture occurs by the propagation of an el

Nutka1998 [239]

Answer:

The theoretical fracture strength of a brittle material is 26465.29 MPa.

Explanation:

Given that,

Length = 0.54 mm

Radius curvature $R=3.4\times10^{-3}\ mm$

Stress = 1050 MPa

We need to calculate the theoretical fracture strength of a brittle material

Using formula of strength

$\sigma_{m}=2\sigma_{0}(\dfrac{l}{\rho_{t}})^{\frac{1}{2}}$

Where, l = length of the crack

$\rho_{t}$ = tip radius of curvature

Put the value into the formula

$\sigma_{m}=2\times1050\times(\dfrac{0.54}{3.4\times10^{-3}})^{\frac{1}{2}}$

$\sigma_{m}=26465.29\ MPa$

Hence, The theoretical fracture strength of a brittle material is 26465.29 MPa.

4 0

3 years ago