All categories

4 years ago

At what rate is 444kg object accelerating if a force of 7,700N is applied to it?

Physics

1 answer:

vlabodo [156]4 years ago

7 0

Answer:

646,800

Explanation:

If you multiply the 7,700N and the 444kg your answer would be 646,800.

You might be interested in

A bus slows with constant acceleration from 24.0 m/s to 16.0 m/s and moves 50.0 m in the process. (a) How much further does it t

navik [9.2K]

Answer:

(a) Bus will traveled further a distance of 40 m

(b) It will take 7.5 sec to stop the bus

Explanation:

We have given initial velocity of the bus u = 24 m/sec

And final velocity v = 16 m/sec

Distance traveled in this process s = 50 m

From third equation of motion we know that $v^2=u^2+2as$

$16^2=24^2+2\times a\times 50$

$a=-3.2m/sec^2$

(a) Now as the bus finally stops so final velocity v = 0 m/sec

So $v^2=u^2+2as$

$0^2=24^2-2\times 3.2\times s$

s= 90 m

So further distance traveled by bus = 90-50 =40 m

(b) Now as the bus finally stops so final velocity v= 0 m/sec

Initial velocity u = 24 m/sec

Acceleration $a=-3.2m/sec^2$

So time $t=\frac{v-u}{a}=\frac{0-24}{-3.2}=7.5sec$

7 0

3 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

3 years ago

Light that has a wavelength of 554 nm is incident on a single slit that is 10 μm wide. Determine the angular location of the fir

VMariaS [17]

66.3 = 3% 9.528 correct 3.176 correct

7 0

4 years ago

Part A Determine the absolute pressure on the bottom of a swimming pool 30.0 m by 8.7 m whose uniform depth is 1.8 m . Express y

Sphinxa [80]

Answer:

17.66 kPa

Explanation:

The volume of water in the swimming pool is the product of its dimensions

V = 30 * 8.7 * 1.8 = 469.8 cubic meters

Let water density $\rho = 1000 kg/m^3$ , and g = 9.81 m/s2 we can calculate the total weight of water in the swimming pool

$W = mg = \rho V g = 1000 * 469.8 * 9.81 = 4608738 N$

The area of the bottom

A = 30 * 8.7 = 261 square meters

Therefore the pressure is its force over unit area

$P = F/A = 4608738 / 261 = 17658 N/m^2$ or 17.66 kPa

7 0

4 years ago

An electric current always produces a magnetic field that is ______ to the field. A. perpendicular B. parallel C. at a 37° angle

Aleks04 [339]

Answer:perpendicular

Explanation:

8 0

3 years ago