A student is performing experiments on a particular substance. Which

A rock with a mass of 8 kg falls straight down from a height of 7 m. What work is done?

fenix001 [56]

F=MA
F=(8 kg)(9.8 m/s)
F= 78.4 N
W=FD
W=(78.4 N)(7 m)
W=548.8 J
How this helps

7 0

4 years ago

A car turns a certain curve of radius 24.98 m with constant linear speed of

Anastaziya [24]

Answer:

3525.19 kg

Explanation:

The computation of the mass of the car is shown below:

As we know that

Fc = m × V^2 ÷ R

m = Fc × R ÷ V^2

Provided that:

Fc = 34.652 kN = 34652 N

R = Radius = 24.98 m

V = speed = 15.67 m/s

So,

m = 34652 × 24.98 ÷ 15.67^2

= 3525.19 kg

7 0

3 years ago

Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

Read 2 more answers

With a mass of 109 kg, Baby Bird is the smallest monoplane ever

OLga [1]

Total resultant velocity=5.11-3.27=1.84m/s

m_1=61.4kg
m_2=109kg
v_1=1.84m/s
v_2=?

$\\ \sf\longmapsto ∆P=P$

$\\ \sf\longmapsto m_1v_1=m_2v_2$

$\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}$

$\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}$

$\\ \sf\longmapsto v_2=112.976/109$

$\\ \sf\longmapsto v_2\approx 1.3m/s$

4 0

3 years ago

If the same pressure is exerted over a greater area will more of less force result? Group of answer choices :

Zigmanuir [339]

Answer:

More force

Explanation:

Pressure and force are related by the equation:

$p=\frac{F}{A}$

where

p is the pressure

F is the force

A is the area

We can re-arrange the equation as

$F=pA$

In this problem, the pressure is kept the same (p' = p) while the area is increased. As we can see from the previous equation, the force applied is directly proportional to the area: therefore, a greater area means also a greater force.

8 0

4 years ago