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Harrizon [31]
2 years ago
12

I need help with 03.01 Playing With Others – President's Challenge Goals

Physics
1 answer:
galben [10]2 years ago
4 0

Answer:

is this like a book orsomthing

Explanation:

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3. What is a manometer used to determine?​
Gemiola [76]

Answer:

It's used to indicate pressure

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2 years ago
Read 2 more answers
A. 2kg<br><br> B. 6kg<br><br> C. 8kg<br><br> D. 14kg
Sati [7]

Answer:

A

Explanation:

We know the force is (16N-12), which is 4N, and we know the acceleration is 2 m/s^2. Meaning, we can solve the formula m = F / a (mass equals force divided by acceleration), and we get 2kg.

7 0
2 years ago
What part of the stem cell provides instructions for building the heart
Monica [59]
It’s DNA that’s a answer
3 0
2 years ago
a car initially at rest move with the constant accerates along straght line read after it's spread increase and finally related
nasty-shy [4]

Answer:

32km per hour

Explanation:

Explanation:

In first case v = a t

==> a t = 40 km p h

Now distance covered S1 + S2 + S3

S1 = 1/2 a t^2 and S3 = 1/2 a t^2

But S2 = 3t * 40 = 120 t km

Hence total distance = at^2 + 120 t

Time taken (total) = t + 3t + t = 5 t

Hence average speed = at^2 + 120 t / 5 t

Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour

8 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
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