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3 years ago

Elements that typically give up electrons

Physics

1 answer:

Bess [88]3 years ago

8 0

Answer:

they are metals

Explanation:

they don't need to gain electrons they need to get rid of them

You might be interested in

Find the current through a 12 ohm resistive circuit when 24 volts is applied

marusya05 [52]

Answer:

Explanation:

Voltage = Current * Resistance

V= I*R

I=V/R

I = 24/12 = 2A.

5 0

4 years ago

This problem, a squid at rest suddenly sees a predator coming toward it and needs to escape. Assume the following:______.

makkiz [27]

Answer:

6.79 m/s

Explanation:

By applying the principle of conservation of momentum.

The total momentum = MV - mv = 0 (since the squid is beginning at rest)

the mass of the squid (M) in absence of water in its cavity = (6.5 - 1.75) kg

= 4.75 kg

speed of the squid (V) = 2.5 m/s

mass of the water expelled (m) = 1.75 kg

speed of the water (v) = ???

∴

4.75 × 2.5 = 1.75 × v

$v = \dfrac{4.75 \times 2.5}{1.75 }$

v = 6.79 m/s

8 0

3 years ago

What will be the weight of a box, mass 10Kg on the surface of a hypothetical planet having mass double of Earth and the radius i

vlada-n [284]

Answer:

W ’= 21.78 kg

Explanation:

The expression for weight is

W = m g

let's look for the acceleration of gravity with the universal law of gravitation

F = G m M / r2

F = m (G M / r2)

without comparing the two equations

g’= G M / r2

in that case M = 2 Mo and r = 3 ro

where mo and ro are the mass and radius of the earth

we substitute

g ’= G 2Mo / (3r₀) 2

G ’= 2/9 G Mo / r₀²

g ’= 2/9 g

the weight of the body on this planet is

W ’= m g’

W ’= m 2/9 g

let's calculate

W ’= 2/9 10 9.8

W ’= 21.78 kg

3 0

3 years ago

A Plane has a takeoff speed of 150 m/s and requires 1500m to reach that speed. Determine the acceleration of the plane and the t

mars1129 [50]

Answer:

The acceleration of the plane and the time required to reach this speed is (a)= 7.5 $m/sec^2$ and time(t) = 20 seconds

Explanation:

Given data Initial velocity $(V_i)$ = 0

Final velocity ( $V_f$ ) = 150 m/second

Distance (d) = 1500 m

We have the formula, $\mathrm{V}_{\mathrm{f}}^{2}=\mathrm{V}_{\mathrm{i}}^{2}+2 \mathrm{ad}$

which gives $150^2$ = 0+2a(1500)

22500 = 3000 a

acceleration (a) = 7.5 $m/s^2$

$\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at}$

150 = 7.5 t

t= 150/7.5 = 20

t = 20 seconds.

5 0

3 years ago

Two objects, one having twice the mass of the other, are initially at rest. Two forces, one twice as big as the other, act on th

Elodia [21]

Note: There is no image to take as a reference, so I'm assuming F2 directed to the right and F1 to the left, and F2=2F1

Answer:

$\displaystyle a=\frac{F}{3M}$

to the right

Explanation:

Net Force

When several forces are applied to a particle or a system of particles, the net force is the sum of them all, considering each force as a vector. As for the second Newton's law, the total force equals the product of the mass by the acceleration of the system:

$\vec F_n=m\cdot \vec a$

If the net force is zero, then the system of particles keeps at rest or at a constant velocity.

The system of particles described in the question consists of two objects of masses m1=M and m2, where

$m_2=2m_1=2M$

Two forces F1=F and F2 act individually on each object in opposite directions and

$F_2=2F_1=2F$

We don't get to see any image to know where the forces are applied to, so we'll assume F2 to the right and F1 to the left.

The net force of the system of particles is

$F_n=2F-F=F$

The mass of the system is

$m_t=m_1+m_2=3M$

Thus, the acceleration of the center of mass of the system is

$\displaystyle a=\frac{F}{3M}$

Since F2 is greater than F1, the direction of the acceleration is to the right.

Note: If the forces were opposite than assumed, the acceleration would be to the left

6 0

3 years ago