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3 years ago

If the truck has a mass of 2,000 kilograms , what's its momentum?(v=35 m/s)

Physics

1 answer:

katen-ka-za [31]3 years ago

8 0

Answer:

$\boxed {\boxed {\sf 70,000 \ kg*m/s}}$

Explanation:

Momentum is the product of mass and velocity.

$p=m*v$

The mass of the truck is 2,000 kilograms and the velocity is 35 meters per second.

$m= 2000 \ kg \\v= 35 \ m/s$

Substitute the values into the formula and multiply.

$p= 2000 \ kg * 35 \ m/s \\p= 70,000 \ kg*m/s$

The truck's momentum is 70,000 kilograms meters per second.

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An example of an action and reaction is sitting on a chair. Your body pushes down on the chair because of the force of what?

vaieri [72.5K]

Your body is pushing down on a chair because it is being attracted by gravity, the reason that your body is not moving down because there is a normal force acting on your body; together the net force of weight (m*g) and normal force is equal to zero

8 0

3 years ago

Consider a motor that exerts a constant torque of 25.0 nâ‹…m to a horizontal platform whose moment of inertia is 50.0 kgâ‹…m2. A

Crazy boy [7]

Answer:

W = 1884J

Explanation:

This question is incomplete. The original question was:

Consider a motor that exerts a constant torque of 25.0 N.m to a horizontal platform whose moment of inertia is 50.0kg.m^2 . Assume that the platform is initially at rest and the torque is applied for 12.0rotations . Neglect friction.

How much work W does the motor do on the platform during this process? Enter your answer in joules to four significant figures.

The amount of work done by the motor is given by:

$W=\Delta K$

$W= 1/2*I*\omega f^2-1/2*I*\omega o^2$

Where I = 50kg.m^2 and ωo = rad/s. We need to calculate ωf.

By using kinematics:

$\omega f^2=\omega o^2+2*\alpha*\theta$

But we don't have the acceleration yet. So, we have to calculate it by making a sum of torque:

$\tau=I*\alpha$

$\alpha=\tau/I$ => $\alpha = 0.5rad/s^2$

Now we can calculate the final velocity:

$\omega f = 8.68rad/s$

Finally, we calculate the total work:

$W= 1/2*I*\omega f^2 = 1883.56J$

Since the question asked to "Enter your answer in joules to four significant figures.":

W = 1884J

3 0

3 years ago

Given the thermochemical equations X2+3Y2⟶2XY3ΔH1=−370 kJ X2+2Z2⟶2XZ2ΔH2=−120 kJ 2Y2+Z2⟶2Y2ZΔH3=−270 kJ Calculate the change in

Alchen [17]

Answer : The change in enthalpy of the reaction is, -310 kJ

According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.

According to this law, the chemical equation can be treated as ordinary algebraic expression and can be added or subtracted to yield the required equation. That means the enthalpy change of the overall reaction is the sum of the enthalpy changes of the intermediate reactions.

The given main reaction is,

$4XY_3+7Z_2\rightarrow 6Y_2Z+4XZ_2$ $\Delta H=?$

The intermediate balanced chemical reaction will be,

(1) $X_2+3Y_2\rightarrow 2XY_3$ $\Delta H_1=-370kJ$

(2) $X_2+2Z_2\rightarrow 2XZ_2$ $\Delta H_2=-120kJ$

(3) $2Y_2+Z_2\rightarrow 2Y_2Z$ $\Delta H_3=-270kJ$

Now we will reverse the reaction 1 and multiply reaction 1 by 2, reaction 2 by 2 and reaction 3 by 3 then adding all the equations, we get :

(1) $4XY_3\rightarrow 2X_2+6Y_2$ $\Delta H_1=2\times (+370kJ)=740kJ$

(2) $2X_2+4Z_2\rightarrow 4XZ_2$ $\Delta H_2=2\times (-120kJ)=-240kJ$

(3) $6Y_2+3Z_2\rightarrow 6Y_2Z$ $\Delta H_3=3\times (-270kJ)=-810kJ$

The expression for enthalpy of formation of $CH_4$ will be,

$\Delta H=\Delta H_1+\Delta H_2+\Delta H_3$

$\Delta H=(+740kJ)+(-240kJ)+(-810kJ)$

$\Delta H=-310kJ$

Therefore, the change in enthalpy of the reaction is, -310 kJ

5 0

3 years ago

Kevin jumps straight up in the air to a height of 1 meter.At the top of his jump, he has potential energy of 1,000 joules.Answer

Llana [10]

Gravitational potential energy can be given by the equation
PE = mgh
where m is the mass,
g is the gravitational constant 9.81 or 10 depending on rounding
and h is the height

well weight is a force equiavlent to
W= m*g

so comparing that to the potential energy equation, divide the potential energy by the height and you will get weight in Newtons

3 0

3 years ago

20 Accelerated motion is represented by a line on a nonlinear distance time graph.

Varvara68 [4.7K]

Curved line

Explanation:

Acceleration of motion is represented by a curved line on a non-linear distance-time graph.

The acceleration of a non-linear motion is depicted using a parabola which is a curve. This implies that the velocity is constantly changing and the distance covered by the body is also changing with equal amount of time.

A plot of this will give a parabola. This can be further established using one of the equations of motion below:

x = u + $\frac{1}{2}$ at ²

This is a quadratic function where:

x is the distance

u is the initial velocity

t is the time

a is acceleration

A quadratic function gives a curved line which is a parabola.

Learn more:

Acceleration brainly.com/question/10932946

#learnwithBrainly

3 0

3 years ago