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3 years ago

What is the force experienced when a 5.0 kg mass is accelerated at 7.5 m/s??

Physics

1 answer:

Fittoniya [83]3 years ago

4 0

Answer:

F = 37.5 N

Explanation:

It is given that,

Mass of the object is 5 kg

Acceleration of the object is 7.5 m/s²

We need to find the force experienced by the object. The force is given by the product of mass and acceleration of an object. So,

F = m

$F=5\times 7.5\\\\F=37.5\ N$

So, the force acting on the object is 37.5 N.

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A 0.25 kg steel ball is tied to the end of a string and then whirled in a vertical circle at a constant speed v. The length of t

TiliK225 [7]

Answer:

Explanation:

Let the tension in the string be T . At the top of the circle , total force acting on them = T + mg . This will provide centripetal force

T + mg = m v² / r

4 + .25 x 9.8 = .25 x v² / .62

6.45 = .25 v² / .62

v² = 16

v = 4 m /s .

7 0

2 years ago

What is the internal energy of 4.50 mol of N2 gas at 253°C? To solve this

Finger [1]

The internal energy of the gas is 49,200 J

Explanation:

The internal energy of a diatomic gas, such as $N_2$ , is given by

$U=\frac{5}{2}nRT$

where

n is the number of moles

R is the gas constant

T is the absolute temperature of the gas

For the gas in this problem, we have:

n = 4.50 (number of moles)

R = 8.31 J/(mol·K) (gas constant)

$T=253^{\circ} + 273 = 526 K$ (absolute temperature)

Substituting, we find:

$U=\frac{5}{2}(4.50)(8.31)(526)=49,200 J$

Learn more about ideal gases:

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3 years ago

Imagine whirling a ball on a string over you head. Suppose the knot holding the ball comes loose and the ball is instantly relea

zaharov [31]

B is the answer to the question

8 0

3 years ago

An open train car moves with speed 18.5 m/s on a flat frictionless railroad track, with no engine pulling the car. It begins to

olga2289 [7]

Answer:

The speed decreases.

Explanation:

This can be explained using the conservation of linear momentum.

Since there is no friction, the initial moment of the train must be equal to its linear moment after it is filled with water.

the initial linear momentum is

$m_{1}v_{1}$

where $m_{1}$ is the initial mass of the train, and $v_{1}$ the initial speed of the train.

And linear momentum after the water filled the train car is

$m_{2}v_{2}$

where $m_{2}$ is mass of the train after the rain, and $v_{2}$ the speed of the train after the rain

<u>the equality must be fulfilled:</u>

$m_{1}v_{1}=m_{2}v_{2}$

We know that if water is added to the train, $m_{2}$ that is the mass after the water is added, is greater than $m_{1}$ which is the mass of the train without the water.

Therefore, in order for the conservation of the linear momentum to be fulfilled: $m_{1}v_{1}=m_{2}v_{2}$

the speed after the water is added ( $v_{2}$ ) must be smaller than the initial train speed ( $v_{1}$ ) . So the speed of the car decreases.

3 0

3 years ago

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

2 years ago