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denis-greek [22]
3 years ago
13

Force of 10N down,10N to the right,and 5N to the left are acting on a ball .it acceleration horizontally to the right.what other

force,if any,is acting on the ball
Physics
1 answer:
denpristay [2]3 years ago
3 0

Answer:

A 10 N force pointing up

Explanation:

If the net acceleration of the object is horizontal pointing to the right, that means that all vertical forces must have canceled out, and the only ones "unbalanced" are the horizontal ones (10 N to the right minus 5 N to the left giving a net force of 5 N to the right).

Since they mentioned only one vertical force pointing down (10 N), there must be another one of same magnitude but pointing in opposite direction (up).

Then there must also be a 10 N force pointing up acting on the object.

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Any projectile always has a constant vertical acceleration of _______.
Mashcka [7]

Answer:

The answer to your question is: C. -9.81 m/s²

Explanation:

A. 9.81 m/s²  acceleration is considered positive when it goes to the center of the earth, so this option is incorrect.

B. 0 m/s²  This option is incorrect because acceleration is 0 for a linear motion without acceleration.

C. -9.81 m/s²  If a projectile goes to the sky, then the acceleration will be negative.

D. It is not constant. Acceleration is constant.

8 0
3 years ago
In being served, a tennis ball is accelerated from rest to a speed of 31.6 m/s. The average power generated during the serve is
joja [24]

Answer:

The force acting on the ball is 92.4 N.

Explanation:

Given that,

Initial speed of the ball, u = 0

Final speed of the ball, v = 31.6 m/s

The average power generated during the serve is 2920 W. Power generated by an object is given by :

P=\dfrac{W}{t}

W is the work done, W = Fd

P=\dfrac{Fd}{t}

Since, v=\dfrac{d}{t}

So,

P=F\times v

F is the force acting on the ball

F=\dfrac{P}{v}\\\\F=\dfrac{2920\ W}{31.6\ m/s}\\\\F = 92.4\ N

So, the force acting on the ball is 92.4 N. Hence, this is the required solution.

7 0
3 years ago
Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did
Ksivusya [100]

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

5 0
2 years ago
It is estimated that uranium is relatively common in the earth's crust, occurring in amounts of 4 g / metric ton. A metric ton i
timama [110]

Answer:

The mass of Uranium present in a 1.2mg sample is 4.8 \cdot 10^{-6}\,mg

Explanation:

The ration between Uranium mass and total sample mass is: \frac{4g}{1000kg} =\frac{4g}{1000000g}=\frac{1}{250000}

For a sample of mass 1.2 mg, the amount of uranium is:

1.2\, mg \cdot \frac{1}{250000}=4.8 \cdot 10^{-6}\,mg

7 0
3 years ago
A certain white dwarf star was once an average star like our Sun. But now it is in the last stage of its evolution and is the si
solmaris [256]

Answer:

4.384 * 10^13

Explanation:

Given the expression :

[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]

Applying the laws of indices

[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]

13.2733 * 10^19 ÷ 3.0276 * 10^6

(13.2733 / 3.0276) * 10^(19 - 6)

4.3840996 * 10^13

= 4.384 * 10^13

6 0
2 years ago
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