Which pair of concurrent forces could produce a resultant force having a magnitude of 10. Newtons?

a police officer parked on the side of the rode clockes a driver at a constant 130km/h (passing the stopped car). if the officer

nignag [31]

Answer

given,

speed of the vehicle = 130 Km/h

officers accelerates at = 12 km/h/s

time taken by the driver to catch the vehicle = ?

at time t second

$130 t = \dfrac{1}{2}at^2$

$260 t - 12 t^2 = 0$

t (12 t - 260 )= 0

12 t = 260

t = 21.67 s

v = u + at

v = 12 × 21.67

v = 260 Km/h

8 0

4 years ago

Some hummingbirds beat their wings more than 50 times per second. A scientist is measuring the time it takes for a hummingbird t

Aliun [14]

Answer:

millisecond

Explanation:

Time is measured in seconds in the SI system. The scientist will measure the time it takes to complete one flap of the wings. In the SI system the use of prefix makes it easier to describe the value. It is as follows

$1\ ns=10^{-9}\ s$

$1\ \mu s=10^{-6}\ s$

$1\ ms=10^{-3}\ s$

$1\ cs=10^{-2}\ s$

Here, the wings flap so fast that the millisecond should be used.

5 0

3 years ago

The distance from earth to the center of our galaxy is about 27,000 ly (1 ly = 1 light year = 9.47 × 10^15 m), as measured by an

kompoz [17]

Answer:

The time is 1206 years.

Explanation:

Given that,

Distance = 27000 ly

$1\ ly =9.47\times10^{15}\ m$

Speed = 0.9990c

We need to calculate the time

Using formula of speed

$v = \dfrac{d}{t}$

$t=\dfrac{d}{v}$

Put the value into the formula

$t =\dfrac{27000\times9.47\times10^{15}}{3\times10^{8}}$

$t=8.523\times10^{11}\ sec$

We need to calculate the time dilation the clock on the spaceship

Using formula of dilation time

$t'=t\sqrt{1-\dfrac{v^2}{c^2}}$

$t'=8.523\times10^{11}\times\sqrt{1-\dfrac{( 0.9990c)^2}{c^2}}$

$t'=8.523\times10^{11}\times\sqrt{1-(0.9990)^2}$

$t'=3.8106\times10^{10}\ sec$

The time convert in years

We know that,

$1\ yr= 3.16\times10^{7}\ sec$

So, the time is

$t'=\dfrac{3.8106\times10^{10}}{3.16\times10^{7}}$

$t'=1206\ years$

Hence, The time is 1206 years.

8 0

3 years ago

You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end

Mariulka [41]

Answer: d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

6 0

4 years ago

Part 1 :

dybincka [34]

Answer:

1. 218.55 N

2. $30.96^{o}$

3. $2.1 m/s^{2}$

Explanation:

Part 1;

Net force $F=mg sin \theta$ where m is mass, g is gravitational force and $\theta$ is the angle of inclination

$F= 46*9.8*sin 29^{o}= 218.55N$

Frictional force, $F_{r}$ is given by

$F_{r} = \mu_{s}mg cos \theta$ where $\mu_{s}$ is the coefficient of static friction

$F_{r} = 0.6*46*9.8 cos 29$

$F_{r}=236.57N$

Since $F_{r}>F$ , therefore, the block doesn’t slip and the frictional force acting is mgh=218.55N

Part 2.

Using the relationship that

Frictional force $F_{s} = \mu_{s} mg cos \theta$

$mg sin \theta =\mu_{s} mg cos \theta$

$\mu_{s}= \frac {sin \theta}{cos \theta}$

$\mu_{s}= tan \theta$

The maximum angle of inclination $\theta = tan^{-1} \mu_{s}$

$\theta = tan^{-1} (0.6)$

$\theta= 30.96^{o}$

Part 3:

Net force on the object is given by

$ma = mg sin 38 - \mu_{k} mg cos 38$ where $\mu_{k}$ is the coefficient of kinetic friction

$a = g ( sin 38 - \mu_{k} cos 38 )$

= 9.8 ( sin 38 - (0.51) cos 38 )

= $2.1m/s^{2}$

7 0

4 years ago