Ask question

Ask question

All categories

3 years ago

15

For a constant force, the impulse could be increased by: A.decreasing the distance. B.keeping the time the force acts constant.

C.increasing the time the force acts. D.none of the other answers.

1 answer:

Andre45 [30]3 years ago

5 0

C. Increasing the time of the force
This is because
Impulse = FΔt

You might be interested in

If matter cannot be created nor destroyed where did the matter in the univerce com from ???????

Fynjy0 [20]

i have no idea

I don't even know how we are here

5 0

2 years ago

The sound level of one person talking at a certain distance from you is 61 dB. If she is joined by 5 more friends, and all of th

Lady_Fox [76]

Answer:

Explanation:

For sound level in decibel scale the relation is

dB = 10 log I / I₀ where I₀ = 10⁻¹² and I is intensity of sound whose decibel scale is to be calculated .

Putting the given values

61 = 10 log I / 10⁻¹²

log I / 10⁻¹² = 6.1

I = 10⁻¹² x 10⁶°¹

$=10^{-5.9}$

intensity of sound of 5 persons

$I=5\times 10^{-5.9}$

$dB=10log\frac{5 X 10^{-5.9}}{10^{-12}}$

= 10log 5 x 10⁶°¹

= 10( 6.1 + log 5 )

= 67.98

sound level will be 67.98 dB .

4 0

2 years ago

a 2.35 water bucket is swung in a full cirlce of radius 0.824 m just fast enough so that the water doesn't fall out the top mean

tiny-mole [99]

Answer:

2.84 m/s

Explanation:

At the top position of the circular trajectory, the normal reaction is zero:

N = 0

So it means that the only force that is providing the centripetal force is the gravitational force (the weight of the bucket). Therefore we have:

$mg = m \frac{v^2}{r}$

where

m is the mass of the water bucket

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

v is the speed of the bucket

r = 0.824 m is the radius of the circle

Solving for v,

$v=\sqrt{gr}=\sqrt{(9.8 m/s^2)(0.824 m)}=2.84 m/s$

4 0

2 years ago

A long string is wrapped around a 6.6-cm-diameter cylinder, initially at rest, that is free to rotate on an axle. The string is

lys-0071 [83]

Answer:

$\omega_f=571.42\ rpm$

Explanation:

It is given that,

Diameter of cylinder, d = 6.6 cm

Radius of cylinder, r = 3.3 cm = 0.033 m

Acceleration of the string, $a=1.5\ m/s^2$

Displacement, d = 1.3 m

The angular acceleration is given by :

$\alpha =\dfrac{a}{r}$

$\alpha =\dfrac{1.5}{0.033}$

$\alpha =45.46\ rad/s^2$

The angular displacement is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{1.3}{0.033}$

$\theta=39.39\ rad$

Using the third equation of rotational kinematics as :

$\omega_f^2-\omega_i^2=2\alpha \theta$

Here, $\omega_i=0$

$\omega_f=\sqrt{2\alpha \theta}$

$\omega_f=\sqrt{2\times 45.46\times 39.39}$

$\omega_f=59.84\ rad/s$

Since, 1 rad/s = 9.54 rpm

So,

$\omega_f=571.42\ rpm$

So, the angular speed of the cylinder is 571.42 rpm. Hence, this is the required solution.

5 0

2 years ago

One hazard of space travel is debris left by previous missions. There are several thousand objects orbiting Earth that are large

Eva8 [605]

Answer:

6666.67 Newtons

Explanation:

The formula F=ma (force is equal to mass multiplied by acceleration) can be used to calculate the answer to this question.

In this case:

mass= 0.1mg= 1*10^-7 kg
velocity= 4.00*10^3 m/s
time= 6.00*10^-8 s

Using velocity and time, acceleration can be calculated as:

a= 6.667*10^10 m/s²

Substituting these values into the formula F=ma, the answer is:

F= (1*10^-7)kg * (6.667*10^10) m/s²
F= 6666.67 Newtons of force

5 0

2 years ago