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3 years ago

. Increasing the pressure on a gas __________ the volume the gas occupies. (Points : 1)

2 answers:

6 0

Decreases. The atoms become more tightly packed causing the volume, and density to shrink, although the mass stays the same.

bulgar [2K]3 years ago

4 0

It decreases the volume the gas occupies as the atom get closer and are tight which causes it to decrease the volume the gas occupies.

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A 3.00 kg cart on a track is pulled by a string so that it accelerates at 2.00 m/s/s. The force of tension in the string is 10.0

icang [17]

Answer:

If the track is horizontal and the string is pulled horizontally, the friction on the cart would be $4.0\; \rm N$ .

Explanation:

Let $m$ denote the mass of this cart, and let $a$ denote the acceleration of this cart.

$m = 3.00\; \rm kg$ .

$a = 2.00\; \rm m \cdot s^{-2}$ .

Apply Newton's Second Law to find the net force on this cart.

$\begin{aligned}\text{Net Force} &= m \cdot a\\ &= 3.00\; \rm kg \times 2.00\; \rm m\cdot s^{-2}\\ &= 6.00\; \rm N\end{aligned}$ .

The following forces act upon this cart:

(downward) gravitational attraction from the earth,
(upward) normal force from the track,
(forward) tension from the string, and
(backward) friction from the track.

Assume that the track is horizontal, and that the string was pulled horizontally. The normal force from the track would exactly balance the downward gravitational attraction from the earth. Hence, the $6.00\; \rm N$ net force on this cart would be equal (in size) to the size of the tension from the string ( $10.0\; \rm N$ ) minus the size of the friction from the track.

In other words:

$\begin{aligned}&\text{Size of Net Force}\\ &= \text{Size of Tension} - \text{size of friction}\end{aligned}$ .

$\begin{aligned}& 6.00\;\rm N = 10.0\; \rm N - (\text{size of friction})\end{aligned}$ .

$\text{size of friction} = 10.0\; \rm N - 6.00\; \rm N = 4.0\; \rm N$ .

4 0

3 years ago

A block is released from rest, at a height h, and allowed to slide down an inclined plane. There is friction on the plane. At th

GREYUIT [131]

Here the block has two work done on it

1. Work done by gravity

2. Work done by friction force

So here it start from height "h" and then again raise to height hA after compressing the spring

So work done by the gravity is given as

$W_g = m_A g(h - h_A)$

Now work done by the friction force is to be calculated by finding total path length because friction force is a non conservative force and its work depends on total path

$W_f = -(\mu m_A g cos\theta)(\frac{h}{sin\theta} + \frac{h_A}{sin\theta})$

$W_f = -\mu m_A g cot\theta(h + h_A)$

Total work done on it

$W = m_A g(h - h_A) - \mu m_A g cot\theta(h + h_A)$

So answer will be

None of these

7 0

3 years ago

Read 2 more answers

3. Un caracol recorre en línea recta una distancia de 10,8 m en 1,5 h.

Hunter-Best [27]

Answer:

NO HABLO ESPANOL

Explanation:

sorry:(

5 0

3 years ago

A 50-kg cart is on an incline of 30 degrees above the horizontal. The cart is at rest. The static coefficient of the slope is 0.

sesenic [268]

The cart is at rest, so it is in equilibrium and there is no net force acting on it. The only forces acting on the cart are its weight (magnitude w), the normal force (mag. n), and the friction force (maximum mag. f ).

In the horizontal direction, we have

n cos(120º) + f cos(30º) = 0

-1/2 n + √3/2 f = 0

n = √3 f

and in the vertical,

n sin(120º) + f sin(30º) + (-w) = 0

n sin(120º) + f sin(30º) = (50 kg) (9.80 m/s²)

√3/2 n + 1/2 f = 490 N

Substitute n = √3 f and solve for f :

√3/2 (√3 f ) + 1/2 f = 490 N

2 f = 490 N

f = 245 N

(pointed up the incline)

4 0

3 years ago

Two parallel conducting plates are separated by 4.00 cm. The electric field strength between the two plates is 5.70×10⁴ V/m.

Nookie1986 [14]

Answer:

2280 V

Explanation:

distance between the plates, d = 4 cm = 0.04 m

Electric field strength, E = 5.7 x 10^4 V/m

The formula for potential difference is given by

V = E x d

V = 5.7 x 10^4 x 0.04

V = 2280 V

thus, the potential difference between the plates is 2280 V.

6 0

4 years ago