Ask question

Ask question

All categories

3 years ago

10

A 60-kilogram man holding a 20-kilogram box rides on a skateboard at a speed of positive 7 meters per second. He throws the box

behind him, giving it a velocity of negative 5 meters per second, with respect to the ground. What is his velocity, in meters per second, after throwing the object?

1 answer:

solniwko [45]3 years ago

4 0

Answer:

$v_1=11\ m/s$

Explanation:

It is given that,

Mass of the man, m₁ = 60 kg

Mass of the box, m₂ = 20 kg

Speed of man+ box, V = +7 m/s

Speed of the box, when it was thrown, v₂ = -5 m/s

To find,

Velocity of the man after throwing the box

Solution,

Let v₁ be the velocity of the man after throwing the box. By using the conservation of momentum, it can be calculated.

$(m_1+m_2)V=m_1v_1+m_2v_2$

$(60+20)\times 7=60\times v_1+20\times (-5)$

$v_1=11\ m/s$

Therefore, the velocity of the man after throwing the object is 11 m/s towards his initial direction.

You might be interested in

A lighthouse is located on a small island 5 km away from the nearest point p on a straight shoreline and its light makes two rev

lisabon 2012 [21]

The distance starting from the point to the lighthouse would be regarded as the hypotenuse.

And also will be the radius of the circle the beam of light is generating at that point.

So get the radius first

r = sqrt (1^2 + 5^2)

r = 5.099 km

find the circumference:

C = 2*pi * 5.099 km

C = 2 * 16.01898094

C = 32.04 km

Then find the speed in km/sec

One revolution: 60/2 = 30 sec per revolution

Speed = 32.04 km/30 sec

S = 1.068 km/sec is the speed of light

7 0

3 years ago

Helphpphphphpphphphpphphpph

adell [148]

Airplane with nose up: The plane's speed through the air is the square root of (80 m/s squared) plus (120 m/s squared. The whole picture is a right triangle, and the plane's speed is the hypotenuse. The angle is the angle whose tangent is (80/120). You can get it from a calculator, a book, a slide rule, or online from the site that rhymes with floogle. The man pulling the load is also a right triangle. The horizontal component is (hypotenuse) times (cosine of the angle). The vertical component is (hypotenuse) times (sine of the same angle). Fill in what you know, look up the sin and cos of 25 degrees and write those in too, and then you can solve for what you have to find.

7 0

3 years ago

The addition of 9.0×105 J is required to convert a block of ice at -10 ∘C to water at 11 ∘C. i need help this is due in less tha

Law Incorporation [45]

The mass of the block of ice is 2.24 kg

Explanation:

The amount of heat needed for the whole process consists of three different amounts of heat:

$Q_1$ : the amount of heat needed to raise the temperature of the block of ice from $-10^{\circ}C$ to $0^{\circ}C$

$Q_2$ : the amount of heat needed to melt the block of ice at melting point

$Q_3$ : the amount of heat needed to raise the temperature of the water from $0^{\circ}C$ to $11^{\circ}C$

The total amount of heat needed can be written as

$Q=Q_1+Q_2+Q_3=mC_i\Delta T_1 + m\lambda_f + mC_w\Delta T_3$

where we have:

$Q=9.0 \cdot 10^5 J$ (total amount of heat required)

m is the mass of the block of ice

$C_i = 2108 J/kg^{\circ}C$ is the specific heat of ice

$\lambda_f=3.34\cdot 10^5 J/kg$ is the latent heat of fusion of ice

$C_w=4186 J/kg^{\circ}C$ is the specific heat capacity of water

$\Delta T_1 = 0-(-10)=10^{\circ}C$ is the change in temperature in the 1st process

$\Delta T_3 = 11-0=11^{\circ}C$ is the change in temperature in the 3rd process

Solving the equation for m, we find the mass of the block of ice:

$m=\frac{Q}{C_i\Delta T_1 + \lambda_f+C_w\Delta T_3}=\frac{9.0\cdot 10^5}{(2108)(10)+3.34\cdot 10^5+(4186)(11)}=2.24 kg$

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

8 0

3 years ago

Could someone help me on #7?

ozzi

i would sat C then the impact is weaker

8 0

3 years ago

The burner on an electric stove has a power output of 2.0 kW. A 710 g stainless steel tea kettle is filled with 20∘C water and p

asambeis [7]

Answer:

The volume of water that was in the kettle is 1170 $cm^{3}$

Explanation:

Given:

Power, P = 2.0 kW = 2000 W, Mass of stainless steel, $m_{s}$ = 710 g = 0.71 kg at temperature of $20^{0} C$

Part A:

If it takes time, t = 3.5 minutes to reach boiling point of water $100^{0} C$ , then from conservation of energy,

Total energy supplied by the burner = Total heat gained by the water and the stainless steel to rise from $20^{0} C$ to $100^{0} C$

i.e. Pt = $m_{s}$ $c_{s}$ (100 - 20 ) + $m_{w}$ $c_{w}$ (100 - 20 )

$m_{w}$ = $\frac{pt - 80m_{s} C_{s} }{80c_{w} } = \frac{2000*3.5*60 - 80*0.71* 450}{80*4200}$

$m_{w}$ = 1.17 kg

where $c_{w}$ = 4200 J/Kgk (specific heat capacity of water), $c_{s}$ = 450 J/Kgk (specific heat capacity of steel)

But volume of water in the the kettle, v = $\frac{mass}{density} = \frac{1.17}{1000}= 1.17 *10^{-3}$ $m^{3}$

∴ v = 1170 $cm^{3}$

4 0

3 years ago