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2 years ago

Acceleration is a change in speed or direction over time. In what two ways does the sled accelerate as it descends?

Physics

1 answer:

Anna007 [38]2 years ago

4 0

Answer:

By force and sloppy surface.

Explanation:

By applying force on the sled and slope of the path are the two ways the sled accelerate as it descends. If there is more friction between sled and the ground then force is required to push the sled to move downward while on the other hand, if the path on which sled moves is sloppy then it will move automatically without the use of force, so these two ways can accelerate the sled.

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A ball is thrown upward with an initial velocity of +9.8 m/s. How high does it reach before it starts descending?

aksik [14]

Hope this helps you.

3 0

3 years ago

A 40 kg girl and an 8.4 kg sled are on the surface of a frozen lake, 15 m apart. By means of a rope, the girl exerts a 5.2 N for

stealth61 [152]

Answer:

(a) $a_s=0.62\frac{m}{s^2}$

(b) $a_s=0.13\frac{m}{s^2}$

Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

$a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}$

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

$a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}$

$x_f=x_0+v_0t \pm \frac{at^2}{2}$

For the girl, we have $x_0=0$ and $v_0=0$ . So:

$x_f_g=\frac{a_gt^2}{2}(1)$

For the sled, we have $v_0=0$ . So:

$x_f_s=x_0_s-\frac{a_st^2}{2}(2)$

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

$x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s$

Now, we solve (1) for $x_f_g$

$x_f_g=\frac{0.13\frac{m}{s^2}(6.32s)^2}{2}\\x_f_g=2.6m\\x_f=2.6m$

5 0

3 years ago

A force of 8,480 N is applied to a cart to accelerate it at a rate of 26.5 m/s2. What is the mass of the cart?

Anni [7]

F=ma
8480=26.5m
m=8480/26.5
m=320
The mass of the cart is 320kg.

8 0

3 years ago

Read 2 more answers

Two resistors, one with 7.00 Ω of resistance and the other with 11.00 Ω of resistance, are connected in series to a 9.00 V batte

IRINA_888 [86]

Answer:

4.5 W

Explanation:

Applying,

P = V²/(R₁+R₂).................. Equation 1

Where P = Power, V = Voltage, R₁ and R₂ = values of the two resistor.

From the question,

Given: V = 9.00 V, R₁ = 7.00 Ω, R₂ = 11.00 Ω

Substitute these values into equation 1

P = 9²/(7+11)

P = 81/(18)

P = 4.5 Watt.

Hence the power dessipated by the two resistors is 4.5 watt

5 0

2 years ago

7. CALCULATE: How much work is done in each of

kifflom [539]

<h3>Answer</h3>

a. 8J

b. 90J

<h3>Notes formula</h3>

W = F × s

W = Work done (J)

F = Force (N)

s = distance (m)

<h3>Known that </h3>

a. F = 4N

s = 2m

b. F = 30N

s = 3m

<h3>Question</h3>

a. W = ..?

b. W = ..?

a. W = F × s

= 4N × 2m = 8J

b. W = F × s

= 30N × 3m = 90J

<em>#Moderators please don't be mean, dont delete my answers just to get approval from your senior or just to get the biggest moderation daily rank.</em>

6 0

2 years ago