A scientist needs to determine the average volume of five water samples collected for an experiment. What is

Will mark the brainliest

AysviL [449]

Answer:

The impulse transferred to the nail is 0.01 kg*m/s.

Explanation:

The impulse (J) transferred to the nail can be found using the following equation:

$J = \Delta p = p_{f} - p_{i}$

Where:

$p_{f}$ : is the final momentum

$p_{i}$ : is the initial momentum

The initial momentum is given by:

$p_{i} = m_{1}v_{1_{i}} + m_{2}v_{2_{i}}$

Where 1 is for the hammer and 2 is for the nail.

Since the hammer is moving down (in the negative direction):

$v_{1_{i}} = -10 m/s$

And because the nail is not moving:

$v_{2_{i}}= 0$

$p_{i} = m_{1}v_{1_{i}} = 0.25 kg*(-10 m/s) = -2.5 kg*m/s$

Now, the final momentum can be found taking into account that the hammer remains in contact with the nail during and after the blow:

$p_{f} = m_{1}v_{1_{f}} + m_{2}v_{2_{f}}$

Since the hammer and the nail are moving in the negative direction:

$v_{1_{f}}$ = $v_{2_{f}}$ = -9.7 m/s

$p_{f} = -9.7 m/s(7 \cdot 10^{-3} kg + 0.25 kg) = -2.49 kg*m/s$

Finally, the impulse is:

$J = p_{f} - p_{i} = - 2.49 kg*m/s + 2.50 kg*m/s = 0.01 kg*m/s$

Therefore, the impulse transferred to the nail is 0.01 kg*m/s.

I hope it helps you!

7 0

3 years ago

A horizontail rod (oriented in the east -west direction) is moved northward at a constant velocity through a magnetic field that

Ede4ka [16]

Answer:

a

Explanation:

6 0

3 years ago

What does relative humidity measure?

FromTheMoon [43]

Answer:

It is D

Explanation:

edge unit test review thing

6 0

3 years ago

Read 2 more answers

Suppose you apply a ﬂame and heat one liter of water, raising its temperature 10°C. If you transfer the same heat energy to two

polet [3.4K]

(a) the water density is $d=1000 kg/m^3$ , and 1 liter corresponds to a volume of $V=1 L=0.001 m^3$ . Therefore we can find the mass of the water in the first case:
$m=dV=(1000 kg/m^3)(0.001 m^3)=1 kg$

The amount of heat supplied to the water to raise its temperature by $\Delta T=10 ^{\circ} C$ is
$Q=m C_s \Delta T$ (1)
where
$C_s = 4.18 kJ/(kg ^{\circ} C}$ is the specific heat capacity of the water.
Using the data, we find
$Q=(1 kg)(4.18 kJ/(kg ^{\circ} C)(10^{\circ} C)=41.8 kJ$

We want to find the increase in temperature if we transfer the same amount of heat Q to 2 liters of water. The mass of 2 liters of water is
$m=dV=(1000 kg/m^3)(0.002 m^3)=2 kg$
And so by re-arranging equation (1) we can calculate the new increase of temperature:
$\Delta T_2 = \frac{Q}{m C_s} = \frac{41.8 kJ}{(2 kg)(4.18 kJ/(kg ^{\circ} C)}=5 ^{\circ} C$

(b) Now we have 3 liters of water. SImilarly to point (a), the mass is now
$m=dV=(1000 kg/m^3)(0.003 m^3)=3 kg$
And so, the increase in temperature if we use the same amount of heat as before is
$\Delta T_3= \frac{Q}{m C_s} = \frac{41.8 kJ}{(3 kg)(4.18 kJ/(kg ^{\circ} C)}=3.3 ^{\circ} C$

5 0

4 years ago

Consider two mass-less springs connected in series. Spring 1 has a spring constant k₁, and spring 2 has a spring constant k₂. A

olga2289 [7]

Explanation:

Below is an attachment containing the solution.

6 0

3 years ago