All categories

3 years ago

At a given moment in time, instantaneous speed can be thought of as the magnitude of instantaneous

Physics

2 answers:

ValentinkaMS [17]3 years ago

7 0

Answer:

the answer is velocity on apex

kaheart [24]3 years ago

5 0

At a given moment in time, the instantaneous speed can be thought of as the magnitude of instantaneous velocity.

Instantaneous speed is the magnitude of the instantaneous velocity, the instantaneous velocity has direction but the instantaneous speed does not have any direction. Hence, the instantaneous speed has the same value as that of the magnitude of the instantaneous velocity. It doesn't have any direction.

You might be interested in

What type of charge do individual hair strands have when standing on end due to static electricity?

aalyn [17]

Answer:

C. I think

Explanation:

C. permanent positive charges

8 0

3 years ago

What interests you the most about Social Psychology? What do you think is one of the most aggravating issues that our society fa

strojnjashka [21]

Looks of us and somesociety faces nowadays?

4 0

3 years ago

The specific heat of soil is 0.20 kcal/kg*C and the specific heat of water is 1.00 kcal/kg*C. This means that if 1 kg of soil an

stiv31 [10]

Answer: The soil will be $4\°C$ warmer than the water.

Explanation:

The heat (thermal energy) absorbed can be found using the following equation:

$Q=m.C.\Delta T$

Where:

$Q$ is the heat

$m$ is the mass of the element

$C$ is the specific heat capacity of the material.

$\Delta T$ is the variation in temperature

In the case of soil we have:

$Q_{soil}=m_{soil}.C_{soil}.\Delta T_{soil}$ (1)

Where:

$Q_{soil}=1 kcal$

$m_{soil}=1 kg$

$C_{soil}=0.2 kcal/kg \°C$

$\Delta T_{soil}$

In the case of water we have:

$Q_{water}=m_{water}.C_{water}.\Delta T_{water}$ (2)

Where:

$Q_{water}=1 kcal$

$m_{water}=1 kg$

$C_{water}=1 kcal/kg \°C$

$\Delta T_{water}$

Isolating $\Delta T$ from both equations:

$\Delta T_{soil}=\frac{Q_{soil}}{m_{soil}.C_{soil}}$ (3)

$\Delta T_{soil}=\frac{1 kcal}{1 kg(0.2 kcal/kg \°C)}$

$\Delta T_{soil}=5\°C$ (4)

$\Delta T_{water}=\frac{Q_{water}}{m_{water}.C_{water}}$ (5)

$\Delta T_{water}=\frac{1 kcal}{1 kg(1 kcal/kg \°C)}$

$\Delta T_{water}=1\°C$ (6)

Comparing (4) and (6) we can find the soil will be $4\°C$ warmer than the water.

8 0

4 years ago

What is the upper block's acceleration if the coefficient of kinetic friction between the block and the table is 0.13?

IgorLugansk [536]

Let us assume that pulley is mass less.

Let the tension produced at both ends of the pulley is T.

We are asked to calculate the acceleration of the block.

Let the masses of two bodies are denoted as $m_{1} \ and\ m_{2}\ respectively$

$Let\ m_{1} =1 kg\ and\ m_{2} =2 kg$

As per this diagram, the body having mass 1 kg is moving downward and the body having mass 2 kg is moving on the surface of the table.

Let the acceleration of each block is a .

For body having mass 1 kg:

The net force acting on 1 kg body will be-

$m_{1} g-T=m_{1} a$ [1]

Here tension in the rope will be vertically upward and weight of the body will be in vertical downward direction.

For body having mass 2 kg:

The coefficient of kinetic friction $[\mu]=0.13$

$Hence\ the\ frictional\ force\ F=\mu N$

$F=\mu m_{2} g$

Hence the net force acting on the body having mass 2 kg-

$T-\mu m_{2} g=m_{2} a$ [2]

Here the tension of the rope is towards right i.e along the direction of motion of the 2 kg block and frictional force is towards left.

Combining 1 and 2 we get-

$m_{1} g-T=m_{1}a$ [1]

$T-\mu m_{2}g= m_{2} a$ [2]

---------------------------------------------------

$[m_{1} -\mu m_{2} ]g=[m_{1} +m_{2} ]a$

$a=\frac{m_{1}-\mu m_{2}} {m_{1}+ m_{2}}*g$

$a=\frac{1-[2*0.13]}{1+2} *9.8\ m/s^2$

$a=\frac{0.74}{3} *9.8\ m/s^2$

$a=2.417 m/s$ [ans]

6 0

3 years ago

Read 2 more answers

4. A 70.0 kg boy and a 45.0 kg girl use an elastic rope while engaged in a tug-of-war on an icy,

marta [7]

Answer:

$a_1 = 1.446m/s^2$

Explanation:

Given

$m_1 = 70.0kg$ -- Mass of the boy

$m_2 = 45.0kg$ -- Mass of the girl

$a_2 = 2.25m/s^2$ -- Acceleration of the girl towards the boy

Required

Determine the acceleration of the boy towards the girl ( $a_1$ )

From the question, we understand that the surface is frictionless. This implies that the system is internal and the relationship between the given and required parameters is:

$m_1 * a_1 = m_2 * a_2$