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3 years ago

5

If a proton is released at the equator and falls toward earth under the influence of gravity, the magnetic force on the proton w

ill be toward the ____ assuming the magnetic field is directed toward the north at this location.

1 answer:

pantera1 [17]3 years ago

3 0

Positive positive side naturally because protons always move to positive side and electron always move to negative side

You might be interested in

A 20 μF capacitor initially charged to 30 μC is discharged through a 1.5 kΩ resistor. Part A How long does it take to reduce the

Natasha_Volkova [10]

Answer:

it will take 36.12 ms to reduce the capacitor's charge to 10 μC

Explanation:

Qi= C×V

then:

Vi = Q/C = 30μ/20μ = 1.5 volts

and:

Vf = Q/C = 10μ/20μ = 0.5 volts

then:

v = v₀e^(–t/τ)

v₀ is the initial voltage on the cap

v is the voltage after time t

R is resistance in ohms,

C is capacitance in farads

t is time in seconds

RC = τ = time constant

τ = 20µ x 1.5k = 30 ms

v = v₀e^(t/τ)

0.5 = 1.5e^(t/30ms)

e^(t/30ms) = 10/3

t/30ms = 1.20397

t = (30ms)(1.20397) = 36.12 ms

Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.

7 0

3 years ago

An airplane accelerates at 16 m/s² from rest. If it lifts off from an aircraft carrier after 6.0 s, how far did it go down the r

Alona [7]

B is the answer
16 x 6.0 = 96
Helping is my pleasure

6 0

2 years ago

A loaded 375 kg toboggan is traveling on smooth horizontal snow at 4.50 m/s when it suddenly comes to a rough region. The region

zmey [24]

Answer:

a) The average friction force exerted on the toboggan is 653.125 newtons, b) The rough region reduced the kinetic energy of the toboggan in 92.889 %, c) The speed of the toboggan is reduced in 73.333 %.

Explanation:

a) Given the existence of non-conservative forces (friction between toboggan and ground), the motion must be modelled by means of the Principle of Energy Conservation and the Work-Energy Theorem, since toboggan decrease its speed (associated with due to the action of friction. Changes in gravitational potential energy can be neglected due to the inclination of the ground. Then:

$K_{1} = K_{2} + W_{f}$

Where:

$K_{1}$ , $K_{2}$ are the initial and final translational kinetic energies of the tobbogan, measured in joules.

$W_{f}$ - Dissipated work due to friction, measured in joules.

By applying definitions of translation kinetic energy and work, the expression described above is now expanded and simplified:

$f\cdot \Delta s = \frac{1}{2}\cdot m \cdot (v_{1}^{2}-v_{2}^{2})$

Where:

$f$ - Friction force, measured in newtons.

$\Delta s$ - Distance travelled by the toboggan in the rough region, measured in meters.

$m$ - Mass of the toboggan, measured in kilograms.

$v_{1}$ , $v_{2}$ - Initial and final speed of the toboggan, measured in meters per second.

The friction force is cleared:

$f = \frac{m\cdot (v_{1}^{2}-v_{2}^{2})}{2\cdot \Delta s}$

If $m = 375\,kg$ , $v_{1} = 4.50\,\frac{m}{s}$ , $v_{2} = 1.20\,\frac{m}{s}$ and $\Delta s = 5.40 \,m$ , then:

$f = \frac{(375\,kg)\cdot \left[\left(4.50\,\frac{m}{s} \right)^{2}-\left(1.20\,\frac{m}{s}\right)^{2}\right]}{2\cdot (5.40\,m)}$

$f = 653.125\,N$

The average friction force exerted on the toboggan is 653.125 newtons.

b) The percentage lost by the kinetic energy of the tobbogan due to friction is given by the following expression, which is expanded and simplified afterwards:

$\% K_{loss} = \frac{K_{1}-K_{2}}{K_{1}}\times 100\,\%$

$\% K_{loss} = \left(1-\frac{K_{2}}{K_{1}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{\frac{1}{2}\cdot m \cdot v_{2}^{2}}{\frac{1}{2}\cdot m \cdot v_{1}^{2}} \right)\times 100\,\%$

$\% K_{loss} = \left(1-\frac{v_{2}^{2}}{v_{1}^{2}} \right)\times 100\,\%$

$\%K_{loss} = \left[1-\left(\frac{v_{2}}{v_{1}}\right)^{2} \right]\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\%K_{loss} = \left[1-\left(\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} }\right)^{2} \right]\times 100\,\%$

$\%K_{loss} = 92.889\,\%$

The rough region reduced the kinetic energy of the toboggan in 92.889 %.

c) The percentage lost by the speed of the tobbogan due to friction is given by the following expression:

$\% v_{loss} = \frac{v_{1}-v_{2}}{v_{1}}\times 100\,\%$

$\% v_{loss} = \left(1-\frac{v_{2}}{v_{1}} \right)\times 100\,\%$

If $v_{1} = 4.50\,\frac{m}{s}$ and $v_{2} = 1.20\,\frac{m}{s}$ , then:

$\% v_{loss} = \left(1-\frac{1.20\,\frac{m}{s} }{4.50\,\frac{m}{s} } \right)\times 100\,\%$

$\%v_{loss} = 73.333\,\%$

The speed of the toboggan is reduced in 73.333 %.

5 0

3 years ago

A force of 1.35 newtons is required to accelerate a book by 1.5 meters/second2 along a frictionless surface. What is the mass of

malfutka [58]

B. seems like the answer, I'm not sure.

3 0

4 years ago

Read 2 more answers

An object of mass 25kg is falling from the height h=10 m. calculate

r-ruslan [8.4K]

Answer:

a=2500J,b=1000K,c=1000J,d=14.142m/s

Explanation:

V²=U²+2gh

V²=0 + 2×10×10=200m/s

a).kinetic energy=(1/2)mv²=(1/2)25×200=2500

potential energy=mgh

p.e=25×10×10=2500J

pe+ke=2500+2500=5KJ

b).mgh=25×10×4=1000J

c). V²=U²+2gh

V²=0+2×10×4

V²=80

kinetic energy=(1/2)mv²

=(1/2)25×80

=1KJ

d). From my first paragraph V²=200

V=√200

V=14.142m/s

6 0

3 years ago