If you increase the force exerted on an object, what will happen to the acceleration?

What traits did Sir William Gilbert have that made him a good scientist?

Gave him good advice to others

4 0

2 years ago

lasie4. A 4 kg object is displaced to the right by a distance of 12 m underthe influence of the following forces: a 17 Nforce pu

Oksanka [162]

The work done by a constant force in a rectilinear motion is given by:

$W=Fd\cos\theta$

where F is the magnitude of the force, d is the distance and θ is the angle between the force and the displacement vector.

In this case we have two forces then we need to add the work done by each of them; for the first force we have a magnitude of 17 N, a displacement of 12 m and and angle of 0° (since both the displacement and the force point right); for the second force we have a magnitude of 36 N, a displacement of 12 m and an angle of 30°. Plugging these values we have that the total work is:

$\begin{gathered} W=(17)(12)\cos0+(36)(12)\cos30 \\ W=578.123 \end{gathered}$

Therefore, the total work done is 578.123 J and the answer is option E

6 0

10 months ago

The plane of a rectangular coil, 3.7 cm by 3.7 cm, is perpendicular to the direction of a uniform magnetic field B. If the coil

gizmo_the_mogwai [7]

Answer:3.77 T/s

Explanation:

Given

Area of cross-section $=3.7\times 3.7 cm^2=13.69 cm^2$

N=no of turns=61

Resistance= $R=8 \Omega$

current (i)=0.04 A

emf induced $=i\times R=0.04\times 8=0.32 V$

emf induced $e=NA\frac{\mathrm{d} B}{\mathrm{d} t}$

$0.32=62\times 13.69\times 10^{-4}\times \frac{\mathrm{d} B}{\mathrm{d} t}$

$\frac{\mathrm{d} B}{\mathrm{d} t}=\frac{0.32\times 10^4}{848.78}=3.77 T/s$

8 0

3 years ago

The vapor pressure of benzene, C6H6, is 40.1 mmHg at 7.6°C. What is its vapor pressure at 60.6°C? The molar heat of vaporization

ANEK [815]

Answer:

The vapor pressure at 60.6°C is 330.89 mmHg

Explanation:

Applying Clausius Clapeyron Equation

$ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]$

Where;

P₂ is the final vapor pressure of benzene = ?

P₁ is the initial vapor pressure of benzene = 40.1 mmHg

T₂ is the final temperature of benzene = 60.6°C = 333.6 K

T₁ is the initial temperature of benzene = 7.6°C = 280.6 K

ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol

R is gas rate = 8.314 J/mol.k

$ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg$