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2 years ago

What is the mass of an object if a 30 N force makes it accelerate at 6 m/s2

Physics

1 answer:

jasenka [17]2 years ago

4 0

Answer:

5 kg

Explanation:

Acceleration = 6 m/s^2

Force = 30 N

Force = mass * acceleration

mass = force / acceleration

mass = 30 / 6

mass = 5 kg

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A ball thrown vertically is caught by the thrower after 5.1s. Find the maximum height the ball reaches.

zimovet [89]

Answer:

h = 31.9 m

Explanation:

Since, the ball took 5.1 s in the air. Hence, the time to reach maximum height will be equal to the half of this value:

t = 5.1 s /2 = 2.55 s

Now, we use 1st equation of motion between the time of throwing and the time of reaching maximum height:

Vf = Vi + gt

where,

Vf = Final Velocity = 0 m/s (since, ball momentarily stops at highest point)

Vi = Initial Velocity = ?

g = - 9.8 m/s² (negative sign for upward motion)

Therefore,

0 m/s = Vi + (-9.8 m/s²)(2.55 s)

Vi = 24.99 m/s

Now, we use second equation of motion for height (h):

h = Vi t + (0.5)gt²

h = (24.99 m/s)(2.55 s) + (0.5)(-9.8 m/s²)(2.55 s)²

h = 63.7 m - 31.8 m

h = 31.9 m

8 0

3 years ago

Determine the resultant moment produced by the load and the weights of the tower crane jibs about point A and about point B. Exp

grigory [225]

Answer:

(M)_A = (M)_b = 76027.5 Nm

Explanation:

Step 1:

- We will first mark each weight from left most to right most.

Point: Weight:

G_3 W_g3 = 6 Mg*g

G_2 W_g2 = 0.5 Mg*g

G_1 W_g1 = 1.5 Mg*g

Load W_L = 2 Mg*g

Step 2:

- Set up a sum of moments about pivot point B, the expression would be as follows:

(M)_b = W_g3 * 7.5 + W_g2*4 - W_g1*9.5 - W_L*12.5

Step 3:

- Plug in the values and solve for (M)_b, as follows:

(M)_b = 9.81*10^3( 6* 7.5 + 0.5*4 - 1.5*9.5 - 2*12.5)

(M)_b = 9.81*10^3 * (7.75)

(M)_b = 76027.5 Nm

Step 4:

- Extend each line of action of force downwards, we can see that value of each force does not change also the moment arms of each individual weight also remains same. Hence. we can say that (M)_A = (M)_b = 76027.5 Nm

5 0

3 years ago

. An object has a position given by ~r(t) = [3.0 m − (4.00 m/s)t]ˆı + [6.0 m − (8.00 m/s2 )t 2 ]ˆ , where all quantities are in

kupik [55]

Answer:

Explanation:

The position is $r(t) = [3.0 \text{ m} - (4.00 \text{ m/s})t]\hat{i} + [6.0 \text{m} - (8.00 \text{ m/s}^2 )t^2 ]\hat{j}$ .

The velocity is the first time-derivative of r(t).

$v(t) = \dfrac{d}{dt}r(t) = -4.00\,\hat{i} -16t\,\hat{j}$

The acceleration is the first time-derivative of the velocity.

$a(t) = \dfrac{d}{dt} v(t) = -16\hat{j}$

Since a(t) does not have the variable t, it is constant. Hence, at any time,

$a = -16\hat{j}$

Its magnitude is 16 m/s².

4 0

3 years ago

Suppose you first walk 12.5 m in a direction 20° west of north and then 24 m in a direction 40° south of west as shown in the fi

Firlakuza [10]

La ce materie vr ca sunt pe tableta

3 0

3 years ago

What is black body radiation? Explain in detail.

tangare [24]

An object that absorbs all radiation falling on it, at all wavelengths, is called a black body. When a black body is at a uniform temperature, its emission has a characteristic frequency distribution that depends on the temperature. Its emission is called black-body radiation

hope it helps

3 0

2 years ago