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3 years ago

How much does a 0.15 kg baseball weigh on earth?

Physics

1 answer:

Gnesinka [82]3 years ago

7 0

1.472 N

to get weight you multiply an object's mass in kilograms with the acceleration of gravity(9.81m/s) :)

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A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose

dusya [7]

Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

5 0

3 years ago

A mass of 5kg starts from rest and pulls down vertically on a string wound around a disk-shaped, massive pulley. The mass of the

Paha777 [63]

Answer:

c. V = 2 m/s

Explanation:

Using the conservation of energy:

$E_i =E_f$

so:

Mgh = $\frac{1}{2}IW^2 +\frac{1}{2}MV^2$

where M is the mass, g the gravity, h the altitude, I the moment of inertia of the pulley, W the angular velocity of the pulley and V the velocity of the mass.

Also we know that:

V = WR

Where R is the radius of the disk, so:

W = V/R

Also, the moment of inertia of the disk is equal to:

I = $\frac{1}{2}MR^2$

I = $\frac{1}{2}(5kg)(2m)^2$

I = 10 kg*m^2

so, we can write the initial equation as:

Mgh = $\frac{1}{2}IV^2/R^2 +\frac{1}{2}MV^2$

Replacing the data:

(5kg)(9.8)(0.3m) = $\frac{1}{2}(10)V^2/(2)^2 +\frac{1}{2}(5kg)V^2$

solving for V:

(5kg)(9.8)(0.3m) = $V^2(\frac{1}{2}(10)1/4 +\frac{1}{2}(5kg))$

V = 2 m/s

8 0

3 years ago

A toy rocket launcher can project a toy rocket at a speed as high as 35.0 m/s.

Anestetic [448]

Answer:

(a) 62.5 m

(b) 7.14 s

Explanation:

initial speed, u = 35 m/s

g = 9.8 m/s^2

(a) Let the rocket raises upto height h and at maximum height the speed is zero.

Use third equation of motion

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

$0^{2}=35^{2}- 2 \times 9.8 \times h$

h = 62.5 m

Thus, the rocket goes upto a height of 62.5 m.

(b) Let the rocket takes time t to reach to maximum height.

By use of first equation of motion

v = u + at

0 = 35 - 9.8 t

t = 3.57 s

The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.

8 0

3 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago

A car drives off a cliff next to a river at a speed of 30 m/s and lands on the bank on theother side. The road above the cliff i

dezoksy [38]

Answer:1.301 s

Explanation:

Given

Initial Velocity(u)=30 m/s

Height of cliff=8.3 m

Time taken to cover 8.3 m

$h=ut+\frac{at^2}{2}$

here Initial vertical velocity is 0

$8.3=\frac{gt^2}{2}$

$t^2=1.69$

$t=1.301 s$

Horizontal distance

$R=u\times t$

$R=30\times 1.301=39.04 m$

7 0

3 years ago