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Oksana_A [137]
4 years ago
11

The rectangular boat shown below has base dimensions 10.0 cm × 8.0 cm. Each cube has a mass of 40 g, and the liquid in the tank

has a density of 1.0 g/mL. How far has the boat sunk into the water?
A. 3.2 cm
B. 4.0 cm
C. 4.8 cm
D. 8.0 cm
Physics
1 answer:
spin [16.1K]4 years ago
3 0

Answer:

A that the answer I think I'm not tryna do u bad thi

Explanation:

but try it

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Can you think of a scenario when the kinetic and gravitational potential energy could both be zero ? Describe or draw how this c
Inga [223]

Both kinetic and gravitational potential energy can become zero at infinite distance from the Earth.

Consider an object  of mass <em>m </em>projected from the surface of the Earth with a velocity <em>v. </em>

The total energy of the body on the surface of the Earth is the sum of its kinetic energy \frac{1}{2} mv^2and gravitational potential energy -\frac{GMm}{R^2}.

here, <em>M</em> is the mass of the Earth, <em>R</em> is the radius of Earth and <em>G</em> is the universal gravitational constant.

The gravitational potential energy of the object is negative since it is in an attractive field, which is the gravitational field of the Earth.

The energy of the object on the surface of the earth is given by,

E_i=\frac{1}{2} mv^2-\frac{GMm}{R^2}

As the object rises upwards, it experiences deceleration due to the gravitational force of the Earth. Its velocity decreases and hence its kinetic energy decreases.

The decrease in kinetic energy is manifested as  an equal increase in potential energy. The potential energy becomes less and less negative as more and more kinetic energy is converted into potential energy.

At a height <em>h</em> from the surface of the Earth, the energy of the object is given by,

E_h=\frac{1}{2} mv_h^2-\frac{GMm}{(R+h)^2}

The velocity v_h is less than <em>v</em>.

When h =∞, the gravitational potential energy increases from a negative value to zero.

If the velocity of projection is adjusted in such a manner that the velocity decreases to zero at infinite distance from the earth, the object's kinetic energy also becomes equal to zero.

Thus, it is possible for both kinetic and potential energies to be zero at infinite distance from the Earth. In this case, kinetic energy decreases from a positive value to zero and the gravitational potential energy increases from  a negative value to zero.


7 0
3 years ago
F an object has a mass of 200 kg and a weight of 1000 N, what is g?
jeka94

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

1,000N = 200g

g = 1000/200

g = 5 m/s square

7 0
3 years ago
A mouse jumps horizontally from a box of height 0.25m.  If the mouse jumps with a speed of 2.1 m/s, how far from the box does th
Sladkaya [172]

The mouse would land 0.47 m away from the box.

3 0
4 years ago
Read 2 more answers
An electron has an initial velocity of (19.0 j + 18.0 k) km/s and a constant acceleration of (3.00 ✕ 1012 m/s2)i in a region in
asambeis [7]

Answer:

E=(-17.08 i +7.2 j -7.6 k )N/C

Explanation:

v= (19.0j+18.k)km/s

a=3.0x10^{13}m/s^2

\beta= 400x10{-6}T

Electron information needed to solve the question:

m_e=9.11x10^{-31}kg

q=-1.6x10{-19}C

F=F_E+F_B=q*(E+Vx \beta)

F=m*a

m*a=q*(E+Vx\beta)

E=\frac{m*a}{q}-(Vx\beta )

E=\frac{9.11x10{-31}kg*3.0x10^{12}m/s^2}{-1.6x10{-19}C}-[(19.0x10^3mj+18.0x10^3m)xi(400x10^{-6}T)]

E=-i17.08N/C-[7.6(-k)+7.2(j)]N/C

E=(-17.08 i +7.2 j -7.6 k )N/C

6 0
3 years ago
Q9 A physics book slides off a horizontal tabletop with a speed of 1.10 m/s. It strikes the floor in 0.350s. ignore air resistan
Rama09 [41]

Answer:

(a) 0.613 m

(b) 0.385 m

(c) vₓ = 1.10 m/s, vᵧ = 3.50 m/s

v = 3.68 m/s², θ = 72.6° below the horizontal

Explanation:

(a)  Take down to be positive.

Given in the y direction:

v₀ = 0 m/s

a = 10 m/s²

t = 0.350 s

Find: Δy

Δy = v₀ t + ½ at²

Δy = (0 m/s) (0.350 s) + ½ (10 m/s²) (0.350 s)²

Δy = 0.613 m

(b) Given in the x direction:

v₀ = 1.10 m/s

a = 0 m/s²

t = 0.350 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (1.10 m/s) (0.350 s) + ½ (0 m/s²) (0.350 s)²

Δx = 0.385 m

(c) Find: vₓ and vᵧ

vₓ = aₓt + v₀ₓ

vₓ = (0 m/s²) (0.350 s) + 1.10 m/s

vₓ = 1.10 m/s

vᵧ = aᵧt + v₀ᵧ

vᵧ = (10 m/s²) (0.350 s) + 0 m/s

vᵧ = 3.50 m/s

The magnitude is:

v² = vₓ² + vᵧ²

v = 3.68 m/s²

The direction is:

θ = atan(vᵧ / vₓ)

θ = 72.6° below the horizontal

3 0
3 years ago
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