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3 years ago

A 5,257 kg rocket blasts off to the moon with an acceleration of 76 m/s ^2 what is the net force on the rocket

1 answer:

7 0

Newton's subsequent law expresses that power is corresponding to what exactly is needed for an object of consistent mass to change its speed. This is equivalent to that item's mass increased by its speed increase.

We use Newtons, kilograms, and meters each second squared as our default units, albeit any proper units for mass (grams, ounces, and so forth) or speed (miles each hour out of every second, millimeters per second², and so on) could unquestionably be utilized also - the estimation is the equivalent notwithstanding.

Hence, the appropriate answer will be 399,532.

Net Force = 399532

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In the sum A→+B→=C→, vector A→ has a magnitude of 13.6 m and is angled 40.2° counterclockwise from the +x direction, and vector

il63 [147K]

Answer:

$|B|=27.00425726m$

$\alpha =210.3781372$ °

Explanation:

Let's use the component method of vector addition:

$A_x=13.6cos(40.2)=10.38762599\\A_y=13.6sin(40.2)=8.778224553\\Cx=13.8cos(20.7+180)=-12.90912763\\Cy=13.8sin(20.7+180)=-4.877952844$

Now, we know:

$C_x=A_x+B_x\\\\C_y=A_y+b_y$

So:

$B_x=C_x-A_x=-23.29675362\\B_y=C_y-A_y=-13.6561774$

Now lets calculate the magnitude of the vector B:

$|B|=\sqrt{(B_x)^{2} +(B_y)^{2} }=27.00425726m$

Finally its angle is given by:

$\alpha =(arctan(\frac{B_y}{B_x}))+180=30.37813438+180=210.3781344$ °

Keep in mind that I added 180 to the angles of C and B to find the real angles measured from the + x axis counter-clock wise.

8 0

3 years ago

A 37 kg object has an applied force of 85N [R] acting on it. The coefficient of

Sliva [168]

Answer:

Explanation:

This is quite tricky! You need to do 2 different equations to solve all the parts of this problem. First is finding the acceleration in one dimension, which has an equation of

F - f = ma

where F is the applied Friction,

f is the frictional force acting against F,

m is the mass of the object, and

a is the acceleration of the object (NOT the velocity!)

This is Newton's Second Law expanded on a bit. The sum of the forces working on an object is equal to the object's mass times its acceleration. We have F, but we need f which is found in the equation

f = μ $F_n$ which is the coefficient of kinetic friction times the weight of the object. Weight is found in the equation

w = mg where m is mass and g is the pull of gravity. Let's start there and work backwards:

w = 37(9.8) to 2 sig figs so

w = 360N. Now fill that in to find f:

f = (.17)(360) to 2 sig figs so

f = 61. Now for the final answer in the original equation way back up at the top:

85 - 61 = 37a and do the subtraction on the left side first:

24 = 37a and then we divide to 2 sig figs to get

a = .65 m/s/s

Since we are moving in a straight line (as opposed to on an angle) the displacement is found in

d = rt which simply says that the distance an object moves is equal to its rate times the time. Therefore,

d = 2.2(3.4) to 2 sig figs so

d = 7.5 m

6 0

3 years ago

When kicking a football, the kicker rotates his leg about the hip joint. If the velocity of the tip of the kicker’s shoe is 35.0

kkurt [141]

Answer:

33.33 rad / s

Explanation:

Linear velocity = 35 m/s

Radius = 1.05 m

The relation between the linear velocity and the angular velocity is given by

Linear velocity = radius × angular velocity

Angular velocity = linear velocity / radius

Angular velocity = 35 / 1.05

= 33.33 rad/ s

6 0

3 years ago

Projectiles that strike objects are good examples of inelastic collisions. A 0.1 kg nail driven by a gas powered nail driver col

Ratling [72]

In an inelastic collision, only momentum is conserved, while energy is not conserved.

1) Velocity of the nail and the block after the collision
This can be found by using the total momentum after the collisions:
$p_f=(m+M)v_f=4.8 kg m/s$
where
m=0.1 kg is the mass of the nail
M=10 kg is the mass of the block of wood
Rearranging the formula, we find $v_f$ , the velocity of the nail and the block after the collision:
$v_f= \frac{p_f}{m+M}= \frac{4.8 kg m/s}{0.1 kg+10 kg}= 0.48 m/s$

2) The velocity of the nail before the collision can be found by using the conservation of momentum. In fact, the total momentum before the collision is given only by the nail (since the block is at rest), and it must be equal to the total momentum after the collision:
$p_i = mv_i = p_f$
Rearranging the formula, we can find $v_i$ , the velocity of the nail before the collision:
$v_i = \frac{p_f}{m}= \frac{4.8 kg m/s}{0.1 kg}=48 m/s$

6 0

3 years ago

Read 2 more answers

A whistle of frequency 564 Hz moves in a circle of radius 71.2 cm at an angular speed of 17.1 rad/s. What are (a) the lowest and

trapecia [35]

Answer:

a) $f'=544.66 \textup{Hz}$