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zhenek [66]
2 years ago
8

An object of mass 0.9 kg is attached to a massless string of length 3 m, and swung with 3 a tangential velocity of 3 m/s. What i

s the force exerted by the string on the object?
a. 32 N
b. 27 N
C. 5 N
d. 2.7 N​
Physics
1 answer:
Ganezh [65]2 years ago
5 0

Answer:

d

Explanation:

you have to divide by the square root and then use the factorial to multiply

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andre [41]

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l.j

Explanation:

7 0
2 years ago
Read 2 more answers
3) Magnets are used to separate steel cans from aluminum cans in recycling plants. How can
Elden [556K]

Answer:

Since steel contains iron (a magnetic metal), the magnets will attract the steel cans since aluminum is not magnetic. This is used to separate the steel cans from the aluminum cans so they can be recycled separately.

4 0
3 years ago
an airplane is flying through a thundercloud at a height of 2000m (this is very dangerous thing to do because of updrafts, turbu
Vedmedyk [2.9K]

Answer

In this question we have given,

Height of plane, h1=2000m

Height at which charge concentration is 40C, h2=3000m

Height at which charge concentration is -40C, h3=1000m

charge concentaration, q1=40C

charge concentaration, q2=-40C

let the charge concentrations at height h2 and h3 as point charges

Now we will first find the electric feild on plane due to positive charge q1=40

E1= k*q1/(h1-h2)..............(1)

Here k=8.98755*10^9N.m^2/C^2

q1=40C

put values of k, q1 , h1 and h2 in equation 1


[tex]E1=(8.98755*10^9)*(40)/(2000-3000)^2\\

E1=[tex]E= 359502+359502\\E=719004 V/mV/m[/tex]

similarly electric feild due to negative charge q2=-40

[tex]E2=(8.98755*10^9)*(-40)/(2000-1000)^2\\

E2=359502V/m

Total electric feild E at the aircraft is given as

E= E1+ E2\\...............(2)

Put values of  E1 and E2 in equation2

\\E=359502+359502\\E= 719004V/m\\

therefore s Total electric feild E at the aircraft is E= 719004V/m

3 0
2 years ago
On a trip to the Colorado Rockies, you notice that when the freeway goes steeply down a hill, there are emergency exits every fe
Zanzabum

Answer:

The maximum speed that the truck can have and still be stopped by the 100m road is the speed that it can go and be stopped at exactly 100m. Since there is no friction, this problem is similar to a projectile problem. You can think of the problem as being a ball tossed into the air except here you know the highest point and you are looking for the initial velocity needed to reach that point. Also, in this problem, because there is an incline, the value of the acceleration due to gravity is not simply g; it is the component of gravity acting parallel to the incline. Since we are working parallel to the plane, also keep in mind that the highest point is given in the problem as 100m. Solving for the initial velocity needed to have the truck stop after 100m, you should find that the maximum velocity the truck can have and be stopped by the road is 18.5 m/s.

Explanation:

4 0
3 years ago
A small airplane takes on 302 l of fuel. If the density of the fuel is 0.821 g/ml, what mass of fuel has the airplane taken on?
ella [17]

To calculate the mass of the fuel, we use the formula

m = V \times  \rho

Here, m is the mass of fuel, V is the volume of the fuel and its value is V =302 \ L =  302 \ L \times \frac{10^{3}m L }{L} = 302 \times 10^{3} \ mL and  \rho is the density and its value of  0.821 g/mL.

Substituting these values in above relation, we get  

m = 302 \times 10^{3} \ mL \times 0.821 g/mL = 247942 g \\\\ m = 247. 94 \ kg

Thus, the mass of the fuel 247 .94 kg.

8 0
3 years ago
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