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oksano4ka [1.4K]
2 years ago
9

Un joven pelotero llamado Saúl en su primer año de ser firmado en grandes ligas, batea varias veces de cuadrangular, el segundo

año batea la cantidad de cuadrangulares del primer año más diez cuadrangulares, el tercer año bateó cinco cuadrangulares menos que el segundo año; si en total de cuadrangulares de los tres años es igual a 135 ¿cuántos cuadrangulares dio este joven en total por cada año?
Ayuda es para mañana :c​
Physics
2 answers:
Brilliant_brown [7]2 years ago
8 0

Answer:

135-15=120

120÷3=40

40+40+(10)+40+(5)=135

Tcecarenko [31]2 years ago
3 0

no se porque y tu si sabes porque yo no

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What were some major reasons for the american television from analog broadcast to digital broadcast?
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Explanation:

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a car is moving 8.80 m/s when it begins to accelerate at 2.45 m/s^2. how much time does it take to trav 138m. please help me (':
Ulleksa [173]

Answer:

7.6 s

Explanation:

Considering kinematics formula for final velocity as

v^{2}=u^{2}+2as

Where v and u are final and initial velocities, a is acceleration and s is distance moved.

Making v the subject then

v=\sqrt{u^{2}+2as}

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

v=\sqrt{8.8^{2}+2*2.45*138}\\v=27.45 m/s

Also, v=u+at and making t the subject of the formula

t=\frac {v-u}{a}

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

t=\frac {27.45-8.8}{2.45}=7.6122448979591\approx 7.6s

Therefore, it needs 7.6 seconds to travel

7 0
3 years ago
Scientists study how the continents move. Why might scientists use a model
tigry1 [53]

Answer:

B. It is too slow to observe directly

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2 years ago
If the load distance of a lever is 30 cm and the effort distance is 60 cm, calculate the amount of effort required to lift a loa
vladimir2022 [97]

Here,

Load distance (Ld) = 30 cm

Effort distance (Ed) = 60 cm

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Effort (E) = ?

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This is a Right answer...

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7 0
3 years ago
A worker lifts a 20.0-kg bucket of concrete from the ground up to the top of a 25.0-m tall building. The bucket is initially at
yuradex [85]

Answer:

Minimum work = 5060 J

Explanation:

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Mass of the bucket (m) = 20.0 kg

Initial speed of the bucket (u) = 0 m/s

Final speed of the bucket (v) = 4.0 m/s

Displacement of the bucket (h) = 25.0 m

Let 'W' be the work done by the worker in lifting the bucket.

So, we know from work-energy theorem that, work done by a force is equal to the change in the mechanical energy of the system.

Change in mechanical energy is equal to the sum of change in potential energy and kinetic energy. Therefore,

\Delta E=\Delta U+\Delta K\\\\\Delta E= mgh+\frac{1}{2}m(v^2-u^2)

Therefore, the work done by the worker in lifting the bucket is given as:

W=\Delta E\\\\W=mgh+\frac{1}{2}m(v^2-u^2)

Now, plug in the values given and solve for 'W'. This gives,

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Therefore, the minimum work that the worker did in lifting the bucket is 5060 J.

7 0
3 years ago
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