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2 years ago

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A 40 kg girl is pushing a 40 kg boy in a 20 kg toy wagon. The wagon is moving at a constant speed of 2.0 m/s. Calculate the comb

ined kinetic energy of the boyWhat is the acceleration of a 50 kg object pushed with a force of 500 newtons?

1 answer:

JulsSmile [24]2 years ago

8 0

Answer:

Combined kinetic energy =120 joules

Acceleration= $10 m/s^2$

Explanation:

Given that the mass of the girl= 40 kg.

The mass of the boy= 40 kg.

The mass of the wagon = 20 kg.

The combined velocity of the boy and the wagon, v= 2.0 m/s.

The kinetic energy of any object having mass m and moving with the speed of v m/s, is

K.E.= $\frac 1 2 mv^2$ .

Here, the combined mass of the boy and the wagon = 40 + 20 = 60 kg.

So, the combined kinetic energy of the boy and the wagon

$=\frac 12 60 \times 2^2= 120$ J

When a force is applied on the object having the mass m, then

F=ma,

where a is the acceleration produced in the object.

Here, F=500 newtons and m= 50 kg.

so, $500=50\times a$

$\Rightarrow a = 500/50=10 m/s^2$ .

Hence, the combined energy is 120 joules and the acceleration is 10 $m/s^2$ .

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Lina20 [59]

Answer:

if the river is 5km wide and is flowing at 4km/hr. eastwards find by scale ... Find either by scale drawing or by calculation (1) the direction in which he must ... He could row his boat directly across the river to point C and then run to B, or he ... A man who can swim at 5km/h in still water swims towards the east to cross arriver.

Explanation:

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2 years ago

Two circular rods, one steel and the other copper, are joined end to end. Each rod is 0.750 m long and 1.50 cm in diameter. The

Eddi Din [679]

Answer:

(a) Steel rod: $1.1 * 10^{-4}$

Copper rod: $1.88 * 10^{-4}$

(b) Steel rod: $8.3 * 10^{-5} m$

Copper rod: $1.41 * 10^{-4} m$

Explanation:

Length of each rod = 0.75 m

Diameter of each rod = 1.50 cm = 0.015 m

Tensile force exerted = 4000 N

(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as

Strain = $\frac{1}{Y} * \frac{F}{A}$

where Y = Young modulus

F = Fore applied

A = Cross sectional area

For the steel rod:

Y = 200 000 000 000 $N/m^{2}$

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

∴ $Strain = \frac{4000}{200000000000 * 0.000177} \\\\Strain = \frac{4000}{35400000}\\ \\Strain = 0.000113 = 1.13 * 10^{-4}$

For the copper rod:

Y = 120 000 000 000 N/m²

F = 4000N

A = $\pi r^{2}$ (r = d/2 = 0.015/2 = 0.0075 m)

=> A = $\pi * (0.0075)^{2}$

=> A = 0.000177 $m^{2}$

$Strain = \frac{4000}{120 000 000 000 * 0.000177} \\\\Strain = \frac{4000}{21240000}\\ \\Strain = = 1.88 * 10^{-4}$

(b) We can find the elongation by multiplying the Strain by the original length of the rods:

Elongation = Strain * Length

For the steel rod:

Elongation = $1.1 * 10^{-4} * 0.75 = 8.3 * 10^{-5} m$

For the copper rod:

Elongation = $1.88 * 10^{-4} * 0.75 = 1.41 * 10^{-4} m$

6 0

2 years ago

A real heat engine operates between temperatures tc and th. during a certain time, an amount qc of heat is released to the cold

tino4ka555 [31]

$q_{c}$ = Heat released to cold reservoir

$q_{h}$ = Heat released to hot reservoir

$W_{max}$ = maximum amount of work

$t_{c}$ = temperature of cold reservoir

$t_{h}$ = temperature of hot reservoir

we know that

$\frac{q_{c}}{q_{h}}=\frac{t_{c}}{t_{h}}$

$q_{h} = (\frac{t_{h}}{t_{c}})q_{c}$ eq-1

maximum work is given as

$W_{max}$ = $q_{h}$ - $q_{c}$

using eq-1

$W_{max}$ = $(\frac{t_{h}}{t_{c}})q_{c}$ - $q_{c}$

6 0

2 years ago

When kicking a football, the kicker rotates his leg about the hip joint. (a) If the velocity of the tip of the kicker’s shoe is

Maurinko [17]

Answer:

Part a)

$\omega = 33.33 rad/s$

Part b)

$F = 500 N$

Part c)

$R_{max} = 40.8 m$

Explanation:

Part a)

As we know that the velocity of the tip of the kicker's shoe is given as

$v = 35 m/s$

also the length of the tip of the shoe from his hip joint is given as

$L = 1.05 m$

now the angular speed is given as

$\omega = \frac{v}{L}$

$\omega = \frac{35}{1.05}$

$\omega = 33.33 rad/s$

Part b)

As we know that force on the ball is given as rate of change in momentum of the ball

so it is given as

$F = \frac{\Delta P}{\Delta t}$

so we have

$F = \frac{m(v_f - v_i)}{\Delta t}$

$F = \frac{0.500(20 - 0)}{20 \times 10^{-3}}$

$F = 500 N$

Part c)

As we know that the formula of range is given as

$R = \frac{v^2 sin(2\theta)}{g}$

now for maximum range we know

$\theta = 45 degree$

$R_{max} = \frac{v^2}{g}$

$R_{max} = \frac{20^2}{9.8}$

$R_{max} = 40.8 m$

6 0

2 years ago

a crate is being lifted into a truck. if it is moved with a 2470n force and 3650 j of work is done , then how far is the crate b

SVEN [57.7K]

Answer:

The crate was being lifted by a height of 1.48 meters.

Explanation:

In an attempt o move a crate;

Force applied = 2470 N

Work done by the force = 3650 J

We know that the work done is defined as the force used to move an object to a distance.

Given the Force used and the work done by that Force, we need to find out the distance the crate was lifted to.

Work done is defined as:

Work = Force*distance covered in the direction of the force

3650 = 2470*distance

distance = 3650/2470

distance = 1.48 meters

4 0

2 years ago