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saul85 [17]
3 years ago
5

A bowler applies 14 N of force to a 4.5 kg bowling ball. Find the ball's acceleration.

Physics
2 answers:
Luba_88 [7]3 years ago
8 0

Answer:

I believe it's  9.5

Explanation:

brilliants [131]3 years ago
4 0

Answer:

F=m(a)  3.111

Explanation:

So to find the acceleration you would have to reverse the problem, a=F/m

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A plastic film moves over two drums. During a 4-s interval the speed of the tape is increased uniformly from v0 = 2ft/s to v1 =
Ratling [72]

Answer:

Question 1)

a) The speed of the drums is increased from 2 ft/s to 4 ft/s in 4 s. From the below kinematic equations the acceleration of the drums can be determined.

v_1 = v_0 + at \\4 = 2 + 4a\\a = 0.5~ft/s^2

This is the linear acceleration of the drums. Since the tape does not slip on the drums, by the rule of rolling without slipping,

v = \omega R\\a = \alpha R

where α is the angular acceleration.

In order to continue this question, the radius of the drums should be given.

Let us denote the radius of the drums as R, the angular acceleration of drum B is

α = 0.5/R.

b) The distance travelled by the drums can be found by the following kinematics formula:

v_1^2 = v_0^2 + 2ax\\4^2 = 2^2 + 2(0.5)x\\x = 12 ft

One revolution is equal to the circumference of the drum. So, total number of revolutions is

x / (2\pi R) = 6/(\pi R)

Question 2)

a) In a rocket propulsion question, the acceleration of the rocket can be found by the following formula:

a = \frac{dv}{dt} = -\frac{v_{fuel}}{m}\frac{dm}{dt} = -\frac{13000}{2600}25 = 125~ft/s^2

b) a = -\frac{v_{fuel}}{m}\frac{dm}{dt} = - \frac{13000}{400}25 = 812.5~ft/s^2

5 0
3 years ago
How are the spiral arms of the milky way detected?
Step2247 [10]
The spiral structure emerges when galactic clusters (open), H II regions and O & B type stars (young stars) are used as tracers. We know this to be true as other pinwheel galaxies exhibit the same patterns across these tracers as in the milky way.
6 0
3 years ago
A 120-kg object and a 420-kg object are separated by 3.00 m At what position (other than an infinitely remote one) can the 51.0-
djverab [1.8K]

Answer:

1.045 m from 120 kg

Explanation:

m1 = 120 kg

m2 = 420 kg

m = 51 kg

d = 3 m

Let m is placed at a distance y from 120 kg so that the net force on 51 kg is zero.

By use of the gravitational force

Force on m due to m1 is equal to the force on m due to m2.

\frac{Gm_{1}m}{y^{2}}=\frac{Gm_{2}m}{\left ( d-y \right )^{2}}

\frac{m_{1}}{y^{2}}=\frac{m_{2}}{\left ( d-y \right )^{2}}

\frac{3-y}{y}=\sqrt{\frac{7}{2}}

3 - y = 1.87 y

3 = 2.87 y

y = 1.045 m

Thus, the net force on 51 kg is zero if it is placed at a distance of 1.045 m from 120 kg.

6 0
3 years ago
A concave spherical mirror has a radius of curvature of magnitude 20.0 cm. (a) Find the location of the image for object distanc
Tems11 [23]

Answer:

Answered

Explanation:

The radius of curvature of the mirror R = 20 cm

then the focal length f = R/2 = 10 cm

(a) From mirror formula

 1/f = 1/di + /1do

then the image distance

    di = fd_o / d_o - f

  = (10)(40) / 40-10

 = 30.76 cm

since the image distance is positive so the image is real

ii) when the object distance d_0=20 cm

 di = 10×20/ 20-10

= 20

Hence, the image must be  real

iii)when the object distance d_0 = 10

di = 10×10 / 10-10 =  ∞ (infinite)

the image will be formed at ∞

here also image will be real but diminished.

7 0
3 years ago
Read 2 more answers
Please answer this. Science 7th grade.
nikdorinn [45]

Answer:

the answer would be 2

Explanation:

it would be 2 because if u look at the diagram the darkest arrow is pointsin towards earth and the moon and when the moon is infront of the sun it cause's an eclispe

6 0
3 years ago
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