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3 years ago

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HELP ME PLEASE If an object is placed under a force of 20 N, it accelerates at a rate of 5 m/s^2. If the force is increased to 5

5 N, what is the new acceleration? a 13.75 kg b 7.25 kg c 21.10 kg d 5.65 kg

2 answers:

Hitman42 [59]3 years ago

6 0

Answer:13.75

Explanation:

lara [203]3 years ago

3 0

Answer:

b

Explanation:

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A toy rocket, launched from the ground, rises vertically with an acceleration of 28 m/s 2 for 9.7 s until its motor stops. Disre

vredina [299]

Answer:

5080.86m

Explanation:

We will divide the problem in parts 1 and 2, and write the equation of accelerated motion with those numbers, taking the upwards direction as positive. For the first part, we have:

$y_1=y_{01}+v_{01}t+\frac{a_1t^2}{2}$

$v_1=v_{01}+a_1t$

We must consider that it's launched from the ground ( $y_{01}=0m$ ) and from rest ( $v_{01}=0m/s$ ), with an upwards acceleration $a_{1}=28m/s^2$ that lasts a time t=9.7s.

We calculate then the height achieved in part 1:

$y_1=(0m)+(0m/s)t+\frac{(28m/s^2)(9.7s)^2}{2}=1317.26m$

And the velocity achieved in part 1:

$v_1=(0m/s)+(28m/s^2)(9.7s)=271.6m/s$

We do the same for part 2, but now we must consider that the initial height is the one achieved in part 1 ( $y_{02}=1317.26m$ ) and its initial velocity is the one achieved in part 1 ( $v_{02}=271.6m/s$ ), now in free fall, which means with a downwards acceleration $a_{2}=-9,8m/s^2$ . For the data we have it's faster to use the formula $v_f^2=v_0^2+2ad$ , where d will be the displacement, or difference between maximum height and starting height of part 2, and the final velocity at maximum height we know must be 0m/s, so we have:

$v_{02}^2+2a_2(y_2-y_{02})=v_2^2=0m/s$

Then, to get $y_2$ , we do:

$2a_2(y_2-y_{02})=-v_{02}^2$

$y_2-y_{02}=-\frac{v_{02}^2}{2a_2}$

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}$

And we substitute the values:

$y_2=y_{02}-\frac{v_{02}^2}{2a_2}=(1317.26m)-\frac{(271.6m/s)^2}{2(-9.8m/s^2)}=5080.86m$

3 0

3 years ago

A 61 kg skater is traveling at 2.5 m/s while carrying a 4.0 kg bowling ball. After he throws the bowling ball forward at twice t

gregori [183]

The final velocity of the skater is 2.34 m/s forward

Explanation:

We can solve this problem by using the law of conservation of momentum. In fact, the total momentum of the system before and after the ball is thrown must be conserved, in absence of external forces.

Before the ball is thrown, the total momentum is:

$p_i = (M+m)u$

where

M = 61 kg is the mass of the skater

m = 4.0 kg is the mass of the ball

u = 2.5 m/s (forward) is the combined velocity of the skater and the ball

After, the ball is thrown at twice the velocity, so the final total momentum is

$p_f = MV+mv$

where

V is the final velocity of the skater

v = 2(2.5) = 5.0 is the final velocity of the ball

Since the total momentum must be conserved, we can write

$p_i = p_f\\(M+m)u = MV+mv\\V=\frac{(M+m)u-mv}{M}=\frac{(61+4.0)(2.5)-(4.0)(5.0)}{61}=2.34 m/s$

So, the skater is moving at 2.34 m/s (forward) after the shot.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

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brainly.com/question/9484203

#LearnwithBrainly

5 0

3 years ago

Adhira loves to ride her bike around the neighborhood. She starts riding 1.2 miles at 30° S of E. Then, she rides another 2.0 mi

zepelin [54]

Answer:

D = 1.8677 miles , θ = 24.28º at South of West

Explanation:

This is an exercise in adding vectors, the easiest way to solve them is to decompose the vectors and add each component algebraically. Let's use trigonometry

first displacement. d = 1.2 miles to 30º south of East

cos ( 360-30) = cos (-30) = x₁ / d

sin (-30) = y₁ / d

x₁ = d cos (-30)

y₁ = d sin (-30)

x₁ = 1.2 cos (-30) = 1,039 miles

y₁ = 1.2 sin (-30) = -0.6 miles

second shift. d = 2.0 miles to 20º West of South

cos (270-20) = x₂ / d

cos (250) = y₂ / d

x₂ = 2.0 cos 250 = -0.684 miles

y₂ = 2.0 sin250 = -1.879 miles

Third displacement. d = 1.6 miles to 30º South of West

cos (180 + 30) = x₃ / d

sin (210) = y₃ / d

x₃ = 1.6 cos 210 = -1.3856 miles

y₃ = 1.6 sin 210 = -0.8 miles

Fourth displacement. d = 2.6 miles to 15º West of North

cos (90 + 15) = x₄ / d

sin (105) = y₄ / d

x₄ = 2.6 cos 105 = -0.6729 miles

y₄ = 2.6 sin 105 = 2,511 miles

having all the components we add

x-axis (West-East direction)

X = x₁ + x₂ + x₃ + x₄

X = 1.039 -0.684 - 1.3846 - 0.6729

X = -1.7025 miles

Y = y₁ + y₂ + y₃ + y₄

Y = -0.6 -1.879 -0.8 +2.511

Y = -0.768

The modulus of this displacement is we use the Pythagorean theorem

D = √ (X² + Y²)

D = √ (1.7025² + 0.768²)

D = 1.8677 miles

let's use trigonometry to find the direction

tan θ = Y / X

θ = tan⁻¹ Y / x

θ = tan⁻¹ (0.768 / 1.7025)

θ = 24.28º

as the two components are negative this angle is in the third quadrant

therefore in cardinal direction form is

θ = 24.28º at South of West

4 0

3 years ago

In one cycle, a heat engine takes in 1000 J of heat from a high-temperature reservoir, releases 600 J of heat to a lower-tempera

Talja [164]

Answer:

η = 40 %

Explanation:

Given that

Qa ,Heat addition= 1000 J

Qr,Heat rejection= 600 J

Work done ,W= 400 J

We know that ,efficiency of a engine given as

$\eta=\dfrac{W(net)}{Q(heat\ addition)}$

Now by putting the values in the above equation ,then we get

$\eta=\dfrac{400}{1000}$

η = 0.4

The efficiency in percentage is given as

η = 0.4 x 100 %

η = 40 %

Therefore the answer will be 40%.

4 0

3 years ago

. A toy rocket has a mass of 350 g at launch. The force it produces

nydimaria [60]

Answer:

the acceleration is reduced by gravity

a = (15 / .35) - [9.8 * sin(65º)]

Explanation:

break the launch vector into two components, vertical and horizontal

Force Net Vertical=-9.8*.350+15cos65 N

force net horizonal=15sin65

initial acceleration= force/mass= (-9.8+15/.350*cos65)j+(15/.350*sin65)i

using i,j vectors..

5 0

3 years ago