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3 years ago

Two people are pulling the same 12 kg box along a flat, frictionless surface. One

Physics

1 answer:

SpyIntel [72]3 years ago

6 0

Answer:

The acceleration is 0.062 m/s^2.

Explanation:

mass = 12 kg

F = 30 N at 20° right

F' = 45 N at 50° left

Let the acceleration is a.

The net force is

F'' = F' cos 50° - F cos 20°

F'' = 45 x 0.643 - 30 x 0.94 = 28.94 - 28.2 = 0.74 N

According to the Newton's second law

0.74 = 12 x a

a = 0.062 m/s^2

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A projectile is launched from the surface of a planet (mass = 15 x 1024 kg, radius = R = 9.6 x 106 m). What minimum launch speed

Serjik [45]

Answer:

The minimum launch speed is 13366.40 m/s.

Explanation:

Given that,

Mass of planet $M=15\times10^{24}\ kg$

Radius $R= 9.6\times10^{6}\ m$

Height = 6R

We need to calculate the speed

Using relation between gravitational force and total energy

$-\dfrac{GMm}{6R+R}=\dfrac{1}{2}mv^2-\dfrac{GMm}{R}$

$\dfrac{}{}mv^2=\dfrac{GMm}{7R}-\dfrac{GMm}{R}$

$\dfrac{1}{2}mv^2=\dfrac{6GMm}{7R}$

$v^2=\dfrac{2\times6GM}{7R}$

Put the value into the formula

$v=\sqrt{\dfrac{2\times6\times6.67\times10^{-11}\times15\times10^{24}}{7\times9.6\times10^{6}}}$

$v=13366.40\ m/s$

Hence, The minimum launch speed is 13366.40 m/s.

6 0

3 years ago

Read 2 more answers

The concrete slab of a basement is 11 m long, 8 m wide, and 0.20 m thick. During the winter, temperatures are nominally 17°C and

mina [271]

Answer:

$\frac{dQ}{dt}= 4312 W$

Explanation:

As we know that base of the slab is given as

$A = 11 \times 8$

$A = 88 m^2$

now we know that rate of heat transfer is given as

$\frac{dQ}{dt} = \frac{kA}{x} (T_2 - T_1)$

here we know that

$k = 1.4 W/m k$

Also we have

$x =0.20$

$\frac{dQ}{dt} = \frac{1.4(88)}{0.20}(17 - 10)$

$\frac{dQ}{dt}= 4312 W$

7 0

3 years ago

Which is true?A)The spin angular momentum vector itself can be measured, and it can be aligned along the measurement axis (z axi

12345 [234]

Answer:

Option A is correct

Explanation:

Direction of spin can be measured because of the behaviours of the particle as it goes through a magnetic field and the size of the spin is dependant on type of particle.

3 0

3 years ago

Two circular loops of wire, each containing a single turn, have the same radius of 3.40 cm and a common center. The planes of th

Shkiper50 [21]

Answer: To solve for the net magnetic field of two circular loops of wire, each containing a single turn

B net= √[( NUI / 2R)^2 × ( NUI / 2R)^2]

Where B net = Net magnetic field

B net = √2 × (NUI/ 2R)

B net =[ √2 × 1 × ( 4 × 3.142 × 10^-7) × (1.8)] ÷ 2 × 0.034

B net = { 1.414 × (0.00000126) × (1.8)}÷ 0.068

B net = 4.72 × 10^-5 T

4 0

3 years ago

Hook's law describes an ideal spring. Many real springs are better described by the restoring force (FSp)s=−kΔs−q(Δs)^3, where q

kvasek [131]

Answer

given,

k = 250 N/m

q = 900 N/m³

(FSp)s=−kΔs−q(Δs)^3

work done = Force x displacement

$W = \int {F. dx}$

limits are x = 0 to x = 0.15 m

work done

$W = \int_0^{0.15} (k x + q x^3)\ dx$

$W = [\dfrac{kx^2}{2}+\dfrac{qx^4}{4}+ C]_0^0.15$

$W = \dfrac{300\times 0.15^2}{2}+\dfrac{900\times 0.15^4}{4}$

W = 3.375 + 0.1139

W = 3.3488 J

b) % cubic term = $\dfrac{0.1139}{3.3488}$

% cubic term = $3.4\ %$

7 0

3 years ago