All categories

3 years ago

A tension of 35.0N is applied downward on a 2.50kg mass. What is the acceleration of the mass?

Physics

2 answers:

Bess [88]3 years ago

7 0

Answer:

$\text{B. }-14.0\:\mathrm{m/s^2}$

Explanation:

From Newton's 2nd Law, we have $\Sigma F=ma$ .

Substituting given values, we get:

$-35.0=2.50a,\\a=\frac{-35.0}{2.50},\\a=\boxed{-14.0\:\mathrm{m/s^2}}$

*Note the negative sign simply implies that the mass is travelling downwards (direction).

Assoli18 [71]3 years ago

3 0

Answer:

D. -23.8 m/s2

Explanation:

You might be interested in

1 poi

tatuchka [14]

Answer:

true

Explanation:

for example assume you are setting in a moving bus and when someone see you from the ground you are in motion but for some who is with you in the bus you are not in motion.

6 0

2 years ago

How much energy from the sun actually reaches the corn answer?

Serhud [2]

The energy from the sun that reaches the corn is about two billionths.

3 0

3 years ago

Read 2 more answers

Water is pumped steadily out of a flooded basement at a speed of 5.4 m/s through a uniform hose of radius 0.83 cm. The hose pass

Gala2k [10]

To solve this problem it is necessary to apply the concepts related to the flow as a function of the volume in a certain time, as well as the potential and kinetic energy that act on the pump and the fluid.

The work done would be defined as

$\Delta W = \Delta PE + \Delta KE$

Where,

PE = Potential Energy

KE = Kinetic Energy

$\Delta W = (\Delta m)gh+\frac{1}{2}(\Delta m)v^2$

Where,

m = Mass

g = Gravitational energy

h = Height

v = Velocity

Considering power as the change of energy as a function of time we will then have to

$P = \frac{\Delta W}{\Delta t}$

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

The rate of mass flow is,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

Where,

$\rho_w$ = Density of water

A = Area of the hose $\rightarrow A=\pi r^2$

The given radius is 0.83cm or $0.83 * 10^{-2}$ m, so the Area would be

$A = \pi (0.83*10^{-2})^2$

$A = 0.0002164m^2$

We have then that,

$\frac{\Delta m}{\Delta t} = \rho_w Av$

$\frac{\Delta m}{\Delta t} = (1000)(0.0002164)(5.4)$

$\frac{\Delta m}{\Delta t} = 1.16856kg/s$

Final the power of the pump would be,

$P = \frac{\Delta m}{\Delta t}(gh+\frac{1}{2}v^2)$

$P = (1.16856)((9.8)(3.5)+\frac{1}{2}5.4^2)$

$P = 57.1192W$

Therefore the power of the pump is 57.11W

6 0

3 years ago

At an altitude of 5000 m the rocket's acceleration has increased to 6.9 m/s2 . What mass of fuel has it burned?

sergey [27]

1) Initial upward acceleration: $6.0 m/s^2$

2) Mass of burned fuel: $0.10\cdot 10^4 kg$

Explanation:

There are two forces acting on the rocket at the beginning:

- The force of gravity, of magnitude $F_g = mg$ , in the downward direction, where

$m=1.9\cdot 10^4 kg$ is the rocket's mass

$g=9.8 m/s^2$ is the acceleration of gravity

- The thrust of the motor, T, in the upward direction, of magnitude

$T=3.0\cdot 10^5 N$

According to Newton's second law of motion, the net force on the rocket must be equal to the product between its mass and its acceleration, so we can write:

$T-mg=ma$ (1)