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aksik [14]
3 years ago
8

Sachi wants to throw a water balloon to knock over a target and win a prize. The target will only fall over if it is hit with a

force of 0.035 N. The water balloon has a mass of 11.4 grams. How fast will Sachi have to make the water balloon accelerate to hit the target with enough force and win the prize?
Physics
2 answers:
Verizon [17]3 years ago
8 0
3.1 the only reason i know this is cause i got it wrong 
Mashutka [201]3 years ago
8 0
Remember that acceleration (a) equals force (F) over mass (m) or a=F/m. In this equation, both the force, 0.035N, and the mass, 11.4 grams, are provided. However, the mass is provided in the wrong unit, as it should be calculated in kilograms. To set up this equation, you will have to convert 11.4 grams into .0114 kilograms then divide .035 N over 0.114 kg to get your answer of 3.07 m/s^2 rounded to 3.1 m/s^2
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If a single circular loop of wire carries a current of 61 A and produces a magnetic field at its center with a magnitude of 1.70
Schach [20]

Answer:

The  radius is  R =  0.22 5 \  m

Explanation:

From the question we are told that

    The current is  I  =  61 \ A

     The  magnetic field is  B  =  1.70 *10^{-4} \  T

Generally the magnetic field produced by a current carrying conductor  is mathematically represented as

        B  =  \frac{\mu_o  *  I}{2 *  R }

=>     R  =  \frac{\mu_o  *  I  }{ 2 *  B }

Here  \mu_o is the permeability of free space with value  \mu_o  =  4\pi * 10^{-7} N/A^2

=>    R  =  \frac{  4\pi * 10^{-7}  *   61  }{ 2 *   1.70 *10^{-4} }

=>  R =  0.22 5 \  m

8 0
3 years ago
A 1000 kg car moves with a constant speed 20.0 m/s
skad [1K]

Answer:

13.33 m/s^2

Explanation:

Velocity^2 then divide that by the radius

6 0
3 years ago
Physical science
attashe74 [19]

Answer:

0.56 km/s

Explanation:

We will define a single system of units for measurement, for this case meters per second [m/s]. That is, we must convert the rest of units such as centimeters per second and kilometers per second to meters per second.

560[\frac{cm}{s}]*(\frac{1m}{100cm} )=5.6[m/s]\\0.56[\frac{km}{s}]*(\frac{1000m}{1km} )=560[m/s]

Therefore the speed of 0.56 [km/s] is the greatest of all

3 0
4 years ago
Which of the following can be computed?
Musya8 [376]
Answer: only the third option. [Vector A] dot [vector B + vector C]

The dot between the vectors mean that the operation to perform is the "scalar product", alson known as "dot product".

This operation is only defined between two vectors, not one scalar and one vector.

When you perform, in the first option, the dot product of any ot the first and the second vectors you get a scalar, then you cannot make the dot product of this result with the third vector.

For the second option, when you perform the dot product of vectar B with vector C you get a scalar, then you cannot make the dot product ot this result with the vector A.

The third option indicates that you sum the vectors B and C, whose result is a vector and later you make the dot product of this resulting vector with the vector A. Operation valid.

The fourth option indicates the dot product of a scalar with the vector A, which we already explained that is not defined.
5 0
3 years ago
A student on a skateboard is moving at a speed of 1.40 m/s at the start of a 2.15 m high and 12.4 m long incline. The total mass
saw5 [17]

Answer:

W = 609.97J

Explanation:

In order to calculate the work done by the student, you take into account that the total work is equal to the change of the kinetic energy of the student, as follow:

W_T=\Delta K\\\\W+W_g-W_f=\frac{1}{2}m(v^2-v_o^2)\\\\W+Mgsin\alpha d-F_fd=\frac{1}{2}m(v^2-v_o^2)         (1)

The work done by the friction force is negative because it is against the motion of the student.

W: work done by the student = ?

Wf: work done by the friction force

Wg: work done by the gravitational force

Ff: total friction force = 41.0N

m: mass of the skateboard = 53.0kg

d: distance traveled by the student

v: final speed of the student = 6.90m/s

vo: initial speed of the student = 1.40m/s

α: angle of the incline

You first calculate the distance d, with the Pythagoras' theorem

d=\sqrt{(2.15m)^2+(12.4m)^2}=12.58m

Furthermore, the angle α is:

\alpha=tan^{-1}(\frac{2.15m}{12.4m})=9.83\°

Then, you solve the equation (1) for W and replace the values of all parameters:

W=\frac{1}{2}m(v^2-v_o^2)+F_fd-Mgsin\alpha d\\\\W=\frac{1}{2}(53.0kg)((6.90m/s)^2-(1.40m/s)^2)+(41.0N)(12.58m)\\-(53.0kg)(9.8m/s^2)sin(9.83\°)(12.58m)\\\\W=609.97J

The work done by the student is 609.97J

6 0
4 years ago
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