Ask question

Ask question

All categories

3241004551 [841]

2 years ago

9

A 2.0-kg projectile moves from its initial position to a point that is displaced 20 m horizontally and 15 m above its initial po

sition. How much work is done by the gravitational force on the projectile

2 answers:

user100 [1]2 years ago

8 0

Answer:

W= - 300 J

Explanation:

Given that

Mass ,m= 2 kg

Horizontal distance ,x= 20 m

Vertical distance ,y= 15 m

As we know that gravity force always acts in the downward direction

F= m g

The displacement of the object is in upward direction

d= -15 m

The work done by a force given as

W= F.d

Now by putting the values

W= - 2 x 10 x 15 ( Take g= 10 m/s²)

W= - 20 x 15

W= - 300 J

Therefore the work done by gravitational force will be -300 J.

Maurinko [17]2 years ago

7 0

Answer:

W = 294 J

Explanation:

given,

mass of the projectile = 2 Kg

horizontal displacement = 20 m

vertical displacement = 15 m

work done by the gravitational force = ?

the work done by gravitational force only account for vertical motion.

force due to gravity = m g

= 2 x 9.8 = 19.6 N

work done is equal force x displacement

W = F x s

W = 19.6 x 15

W = 294 J

You might be interested in

Floor lamps usually have a base with large inertia, while the long body and top have much less inertia. Part A If you want to sh

Novay_Z [31]

Answer:

When a an object is been rotated its resistance capacity to that rotational force is know as rotational inertia and this mathematically given as

$I = mr^2$

Where m is the mass

r is the rotation radius

For the spinning of the lamp as a baton to work the location of the center of mass of the floor lamp needs to be located

This is more likely to be located closer to base of the lamp as compared to the top, so success of spinning a floor lamp like a baton is highly likely if the lamp is grabbed closer to the base because that is where the position of its center of mass is likely to be.

Explanation:

5 0

2 years ago

A na+ ion moves from inside a cell, where the electric potential is -72 mv, to outside the cell, where the potential is 0 v. wha

Vlada [557]

The change in electric potential energy of the ion is equal to the charge multiplied by the voltage difference:
$\Delta U = q \Delta V = q (V_f - V_i)$
where the charge q of the na+ ion is equal to one positive charge, so it's equal to the proton charge: $q=1.6 \cdot 10^{-19} C$ , and Vf and Vi are the final and initial voltages.

Substituting the numbers, we find:
$\Delta U = (1.6 \cdot 10^{-19}C)(0 V-(-0.072 V))=1.15 \cdot 10^{-20} J$

7 0

2 years ago

Read 2 more answers

A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri

vekshin1

Electric charge on the plastic cube: $1.3\cdot 10^{-7}C$

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

$V(r)=\frac{kQ}{r}$

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

$Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V$

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

$qV' = \frac{1}{2}mv^2$

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

$q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C$

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0

2 years ago

A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it

Degger [83]

Answer:

a) $\omega_1=8.168\,rad.s^{-1}$

b) $n_1=7.735 \,rev$

c) $\alpha_1 =0.6864\,rad.s^{-2}$

d) $\alpha_2=4.1454\,rad.s^{-2}$

e) $t_2=1.061\,s$

Explanation:

Given that:

initial speed of turntable, $N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}$

full speed of rotation, $N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}$

time taken to reach full speed from rest, $t_1=11.9\,s$

final speed after the change, $N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}$

no. of revolutions made to reach the new final speed, $n_2=11\,rev$

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

$\omega=\frac{2\pi.N}{60}\, rad.s^{-1}$

where N = angular speed in rpm.

putting the respective values from case 1 we've

$\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}$

$\omega_1=8.168\,rad.s^{-1}$

(c)

using the equation of motion:

$\omega_1=\omega_0+\alpha . t_1$

here α is the angular acceleration

$78=0+\alpha_1\times 11.9$

$\alpha_1 = \frac{8.168 }{11.9}$

$\alpha_1 =0.6864\,rad.s^{-2}$

(b)

using the equation of motion:

$\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1$

$8.168^2=0^2+2\times 0.6864\times n_1$

$n_1=48.6003\,rad$

$n_1=\frac{48.6003}{2\pi}$

$n_1=7.735\, rev$

(d)

using equation of motion:

$\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2$

$12.5664^2=8.168^2+2\alpha_2\times 11$

$\alpha_2=4.1454\,rad.s^{-2}$

(e)

using the equation of motion:

$\omega_2=\omega_1+\alpha_2 . t_2$

$12.5664=8.168+4.1454\times t_2$

$t_2=1.061\,s$

4 0

2 years ago

A 2100 kg car starts from rest and accelerates at a rate of 2.6 m/s2 for 4.0 s. Assume that the force acting to accelerate the c

Reika [66]

When the system is experiencing a uniformly accelerated motion, there are a set of equations to work from. In this case, work is energy which consist solely of kinetic energy. That is, 1/2*m*v2. First, let's find the final velocity.

a = (vf - v0)/t
2.6 = (vf - 0)/4
vf = 10.4 m/s

Then W = 1/2*(2100 kg)*(10.4 m/s)2
W = 113568 J = 113.57 kJ

8 0

2 years ago