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3 years ago

. A toy rocket has a mass of 350 g at launch. The force it produces

Physics

1 answer:

nydimaria [60]3 years ago

5 0

Answer:

the acceleration is reduced by gravity

a = (15 / .35) - [9.8 * sin(65º)]

Explanation:

break the launch vector into two components, vertical and horizontal

Force Net Vertical=-9.8*.350+15cos65 N

force net horizonal=15sin65

initial acceleration= force/mass= (-9.8+15/.350*cos65)j+(15/.350*sin65)i

using i,j vectors..

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A tow truck exerts a net horizontal force of 1050 N on a 760-kg car. What is the acceleration of the car during this time?

vovikov84 [41]

We can solve the problem by using Newton's second law of motion:
$F=ma$
where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object

In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:
$a= \frac{F}{m}= \frac{1050 N}{760 kg}=1.4 m/s^2$

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>B) 1.4 m/s2 horizontally.</span>

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3 years ago

3) How far will 20 N of force stretch a spring with a spring constant of 140 N/m?

muminat

Answer:

7 meters, 2.8 meters

Explanation:

work done (nm) = force (n) * distance (m)

140= 20 * m

140/20 = m

m=7 meters

140= 50 * m

140/50 = m

m= 2.8 meters

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2 years ago

Ways in which a teacher plays a role in the literacy development of the learners

RideAnS [48]

Answer:

encourage all attempts at reading, writing, and speaking

Explanation:

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1 year ago

A car with mass 950 kg and a speed of 16 m/s approaches an intersection. A 1300 kg minivan traveling at 21 m/s is heading for th

Alex73 [517]

Answer:

$V_f = 13.8863 \angle 60.89\°$

Explanation:

Our values are,

$m_1 = 950Kg\\v_1 = 16m/s \\m_2 =1300Kg\\v_2 = 21m/s$

We have all the values to apply the law of linear momentum, however, it is necessary to define the two lines in which the study will be carried out. Being an intersection the vehicle of mass m_1 approaches through the X axis, while the vehicle of mass m_2 approaches by the y axis. In the collision equation on the X axis, we despise the velocity of object 2, since it does not come in this direction.

$m_1v_1=(m_1+m_2)v_fcos\theta$

For the particular case on the Y axis, we do the same with the speed of object 1.

$m_2v_2=(m_1+m_2)v_fsin\theta$

By taking a final velocity as a component, we can obtain the angle between the two by relating the equations through the tangent

$Tan\theta = \frac{m_2v_2}{m_1v_1}\\Tan\theta = \frac{1300*21}{950*16}\\\theta = tan^{-1}(1.7960)\\\theta = 60.89\°$

Replacing in any of the two functions, given above, we will find the final speed after the collision,

$(950)(16)=(950+1300)V_fcos(60.89)$

$V_f= \frac{(950)(16)}{(950+1300)cos(60.89)}$

$V_f = 13.8863 \angle 60.89\°$

8 0

3 years ago

A flow of electric charge in a wire normally requires a _________.

just olya [345]

Flow of electric charge in a wire requires " ELECTRONS "

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3 years ago