What is the value of acceleration as a body moves with an uniform speed

What do borrowers use to secure a mortgage loan? Check all that apply.

Musya8 [376]

Answer:

a house or a land is used to secure a mortgage loan

5 0

2 years ago

A 560-g squirrel with a surface area of 930cm2 falls from a 5.0-m tree to the ground. Estimate its terminal velocity. (Use a dra

bija089 [108]

Explanation:

Terminal velocity is given by:

$v_t=\sqrt{\frac{2mg}{\rho C_dA}}$

Here, m is the mass of the falling object, g is the gravitational acceleration, $C_d$ is the drag coefficient, $\rho$ is the fluid density through which the object is falling, and A is the projected area of the object. in this case the projected area is given by:

$A=\frac{A_s}{2}=\frac{930cm^2}{2}=465cm^2\\465cm^2*\frac{1m^2}{10^4cm^2}=0.0465m^2\\560g*\frac{1kg}{10^3g}=0.56kg$

Recall that drag coefficient for a horizontal skydiver is equal to 1 and air density is $1.28\frac{kg}{m^3}$ .

$v_t=\sqrt{\frac{2(0.56kg)(9.8\frac{m}{s^2})}{(1.28\frac{kg}{m^3}(1)(0.0465m^2)}}\\v_t=13.58\frac{m}{s}$

Without drag contribution the motion of the person is an uniformly accelerated motion, thus:

$v_f^2=v_o^2+2gh\\v_f=\sqrt{2gh}\\v_f=\sqrt{2(9.8\frac{m}{s^2})(5m)}\\v_f=9.9\frac{m}{s}$

7 0

2 years ago

If izzy mass is 0.3kg he applide 657.9n force what will be the accelration

Sholpan [36]

Answer:

The acceleration of the body, a = 2193 m/s²

Explanation:

Given,

The mass of the body, m = 0.3 kg

The force acting on the body, F = 657.9 N

The force acting on an object is proportional to the product of mass and acceleration of the body.

F = m x a

Therefore, the acceleration of the body is

a = F / m

= 657.9 N / 0.3 kg

= 2193 m/s²

Hence, the acceleration of the body, a = 2193 m/s²

4 0

3 years ago

You know your mass is 62 kg but when you stand on a bathroom scale in an elevator it says your mass is 77 kg what is the acceler

lbvjy [14]

In a stationary situation, the weight of person is
$W=mg=(62 kg)(9.81 m/s^2) = 608.2 N$
This is the weight "felt" by the scale, which is basically the normal reaction applied by the scale on the person, and which uses the value of g (9.81) as reference to convert the weight (602.8 N) into a mass (62 kg).

When the person is in the elevator, the scale says 77 kg. The scale is still using the same value of conversion (9.81), so the apparent weight "felt" by the scale is
$W' = m'g=(77 kg)(9.81 m/s^2)=755.4 N$
This is the normal reaction applied by the scale on the person, and which is directed upward. Besides this force, there is still the weight W of the person, acting downward. So, if we use Newton's second law:
$\sum F = ma$
$W-W'=ma$
where a is the acceleration of the elevator. If we solve for a, we find
$a= \frac{W-W'}{m}= \frac{608.2N-755.4N}{62 kg}=-2.37 m/s^2$
The negative sign means the acceleration is in the opposite direction of g (which we take positive), so it means the elevator is going upward.

4 0

2 years ago

Calculate the force of gravity between a comet with a mass of 500kg and a small asteroid with a mass of 20kg that is separated b

givi [52]

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