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3 years ago

a bowling ball is pushed with a force of 22.0 n and acclerates at 5.5m/s2. what mass of of the bwling ball

Physics

2 answers:

Goshia [24]3 years ago

8 0

Answer:

The mass of the bowling ball is 4 kg.

Explanation:

Given that,

Force = 22.0 N

Acceleration = 5.5 m/s²

We need to calculate the mass of the bowling ball

Using newton's second law

$F = ma$

Where, F = force

m = mass of ball

a = acceleration

Put the value into the formula

$22.0=m\times5.5$

$m =\dfrac{22.0}{5.5}$

$m=4\ kg$

Hence, The mass of the bowling ball is 4 kg.

inn [45]3 years ago

4 0

A force can cause a resting object to move, or it can accelerate a moving object by changing the objects speed or direction

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Look at the vector r plotted below

solong [7]

Answer: option A. r = 3x+2y

Explanation:

Vector r is plotted on the graph. On x-axis each small division corresponds to 1 unit. Similarly on y-axis, each small division corresponds to one unit.

The vector is the resultant of addition of its x and y components. we would draw perpendicular to the x-axis and y-axis from the head of vector r.

On x-axis, $r_x$ = +3 units

on y-axis, $r_y$ = +2 units

$r = r_x+r_y$

Hence, vector r can be written as: r = 3x + 2y . Correct option is A.

7 0

4 years ago

Work and Energy

Sergeeva-Olga [200]

a) $W_1 = 2332 J$ , $W_2= 2332 J$

The work done by the student in each trial is equal to the gravitational potential energy gained by the student:

$W=mg\Delta h$

where

m = 68 kg is the mass of the student

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h$ is the gain in height of the student

For the first student, $\Delta h = 3.5 m$ , so the work done is

$W_1 = (68)(9.8)(3.5)=2332 J$

The second student runs up to the same height (3.5 m), so the work done by the second student is the same:

$W_2 = (68)(9.8)(3.5)=2332 J$

2) $P_1 = 204.6 W$ , $P_2 = 274.4 W$

The power exerted by each student is given by

$P=\frac{W}{t}$

where

W is the work done

t is the time taken

For the first student, $W_1 = 2332 J$ and $t=11.4 s$ , so the power exerted is

$P_1 = \frac{W_1}{t_1}=\frac{2332 J}{11.4 s}=204.6 W$

For the second student, $W_2 = 2332 J$ and $t=8.5 s$ , so the power exerted is

$P_2 = \frac{W_2}{t_2}=\frac{2332 J}{8.5 s}=274.4 W$

5 0

3 years ago

You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric

marysya [2.9K]

Answer:

The Electric flux will be $0.42\times10^6\ \rm N.m^2/C$

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

$E=1.36\times10^6\ \rm N/C$

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

$\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\$

Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let $\phi$ be the flux of the Electric Field coming out\passing through it which is given by

$\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C$

It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is $0.42\times10^6\ \rm N.m^2/C$

8 0

3 years ago

Which one i need help

Lostsunrise [7]

Answer:

Explanation:

7 0

3 years ago

A wheelbarrow is experiencing two forces. One is pushing 5 N to the east and the other 5 N in the exact opposite direction. What

Greeley [361]

D) It isn't accelerating
This is because if both sides are pushing with the exact same force, than one force is not overpowering the other, therefor the wheelbarrow would not be moving.

6 0

3 years ago

Read 2 more answers