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3 years ago

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If the force causing a mass to accelerate is multiplied by a number while keeping the mass the same, what happens to the acceler

ation of the object?

1 answer:

klasskru [66]3 years ago

3 0

Answer: If mass remains constant and you double the force you will double the acceleration.

You might be interested in

What is the matching nitrogen base sequence for the gene below?

grigory [225]

Answer:

A

Explanation:

T-G-T is the answer .

the matching Nitrogen base sequence for gene ATC is A

6 0

3 years ago

Given two metal balls (that are identical) with charges LaTeX: q_1q 1and LaTeX: q_2q 2. We find a repulsive force one exerts on

Romashka [77]

Answer:

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

Explanation:

According to Coulomb's law, the magnitude of force between two point object having change $q_1$ and $q_2$ and by a dicstanced is

$F_c=\frac{1}{4\pi\spsilon_0}\frac{q_1q_2}{d^2}-\;\cdots(i)$

Where, $\epsilon_0$ is the permitivity of free space and

$\frac{1}{4\pi\spsilon_0}=9\times10^9$ in SI unit.

Before dcollision:

Charges on both the sphere are $q_1$ and $q_2$ , d=20cm=0.2m, and $F_c=1.35\times10^{-4}$ N

So, from equation (i)

$1.35\times10^{-4}=9\times10^9\frac{q_1q_2}{(0.2)^2}$

$\Rightarrow q_1q_2=6\times10^{-16}\;\cdots(ii)$

After dcollision: Each ephere have same charge, as at the time of collision there was contach and due to this charge get redistributed which made the charge density equal for both the sphere t. So, both have equal amount of charhe as both are identical.

Charges on both the sphere are mean of total charge, i.e

$\frac{q_1+q_2}{2}$

d=20cm=0.2m, and $F_c=1.406\times10^{-4}$ N

So, from equation (i)

$1.406\times10^{-4}=9\times10^9\frac{\left(\frac{q_1+q_2}{2}\right)^2}{(0.2)^2}$

$\Rightarrow (q_1+q_2)^2=2.50\times10^{-15}$

$\Rightarrow q_1+q_2=\pm5\times 10^{-8}$

As given that the force is repulsive, so both the sphere have the same nature of charge, either positive or negative, so, here take the magnitude of the charge.

$\Rightarrow q_1+q_2=5\times 10^{-8}\;\cdots(iii)$

$\Rightarrow q_1=5\times 10^{-8}-q_2$

The equation (ii) become:

$(5\times 10^{-8}-q_2)q_2=6\times10^{-16}$

$\Rightarrow -(q_2)^2+5\times 10^{-8}q_2-6\times10^{-16}=0$

$\Rightarrow q_2=3\times10^{-8}, 2\times10^{-8}$

From equation (iii)

$q_1=2\times10^{-8}, 3\times10^{-8}$

So, the magnitude of initial charges on both the sphere are $3\times10^{-8}$ Coulombs $=0.03 \mu C$ and $2\times10^{-8}$ Colombs or $0.02 \mu C$ .

Considerion the nature of charges too,

$q_1=\pm0.03 \mu C$ and $q_2=\pm0.02 \mu C.$

4 0

3 years ago

Use the circuit diagram to determine the voltage drop across the 8 Ω resistor.

Rainbow [258]

First we need to find the current flowing in the circuit. The three resistors are in series, so the equivalent resistance of the circuit is the sum of the three resistances:
$R_{eq} = R_1 + R_2 + R_3 = 10.0 \Omega + 8.0 \Omega +6.0 \Omega = 24.0 \Omega$

Then we can apply Ohm's law to the whole circuit, to find the current flowing:
$I= \frac{V}{R_{eq}}= \frac{12 V}{24.0 \Omega}=0.5 A$

And now we can apply Ohm's law to the resistor of $8.0 \Omega$ to find the voltage drop across it:
$V_{8\Omega} =IR=(0.5 A)(8.0 \Omega)=4 V$

8 0

3 years ago

Read 2 more answers

A string is 1.6 m long. One side of the string is attached to a force sensor and the other side is attached to a ball with a mas

Sergio039 [100]

Complete Question

The diagram for this question is shown on the first uploaded image

Answer:

The distance traveled in horizontal direction is $D = 1.38 m$

Explanation:

From the question we are told that

The length of the string is $L = 1.6 \ m$

The mass of the ball is $m = 200 g = \frac{200}{1000} = 0.2 \ kg$

The height of ball is $h = 1.5 \ m$

Generally the work energy theorem can be mathematically represented as

$PE = KE$

Where PE is the loss in potential energy which is mathematically represented as

$PE =mgh$

Where h is the difference height of ball at A and at B which is mathematically represented as

$h = y_A - y_B$

So $PE =mg(y_A - y_B)$

While KE is the gain in kinetic energy which is mathematically represented as

$KE = \frac{1}{2 } (v_b ^2 - 0)$

Where $v_b$ is the velocity of the of the ball

Therefore we have from above that

$PE =KE \equiv mg (y_A - y_B) = \frac{1}{2} m (v_b ^2 - 0)$

Making $v_b$ the subject we have

$v_b = \sqrt{2g (y_A - y_B)}$

substituting values

$v_b = \sqrt{2g (1.5 - 0.40)}$

$v_b = 4.6 \ m/s$

Considering velocity of the ball when it hits the floor in terms of its vertical and horizontal component we have

$v_x = 4.6 m/s \ while \ v_y = 0 m/s$

The time taken to travel vertically from the point the ball broke loose can be obtained using the equation of motion

$s = v_y t - \frac{1}{2} g t^2$

Where s is distance traveled vertically which given in the diagram as $s = -0.4$

The negative sign is because it is moving downward

Substituting values

$-0.4 = 0 -\frac{1}{2} * 9.8 * t^2$

solving for t we have

$t = 0.3 \ sec$

Now the distance traveled on the horizontal is mathematically evaluated as

$D = v_b * t$

$D = 4.6 * 0.3$

$D = 1.38 m$

8 0

3 years ago

To get up on the roof, a person (mass 70.0kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad

Julli [10]

The magnitude of the forces acting at the top are;

$\mathbf{F_{Top, \ x}}$ = 132.95 N

$\mathbf{F_{Top, \ y}}$ = 0

The magnitude of the forces acting at the bottom are;

$\mathbf{F_{Bottom, \ x}}$ = $\mathbf{ F_f}$ = -132.95 N

$\mathbf{F_{Bottom, \ y}}$ = 784.8 N

The known parameters in the question are;

The mass of the person, m₁ = 70.0 kg

The length of the ladder, l = 6.00 m

The mass of the ladder, m₂ = 10.0 kg

The distance of the base of the ladder from the house, d = 2.00 m

The point on the roof the ladder rests = A frictionless plastic rain gutter

The location of the center of mass of the ladder, C.M. = 2 m from the bottom of the ladder

The location of the point the person is standing = 3 meters from the bottom

g = The acceleration due to gravity ≈ 9.81 m/s²

The required parameters are;

The magnitudes of the forces on the ladder at the top and bottom

The strategy to be used;

Find the angle of inclination of the ladder, θ

At equilibrium, the sum of the moments about a point is zero

The angle of inclination of the ladder, θ = arccos(2/6) ≈ 70.53 °C

Taking moment about the point of contact of the ladder with the ground, <em>B </em>gives;

$\sum M_B$ = 0

Therefore;

$\sum M_{BCW}$ = $\sum M_{BCCW}$

Where;

$\sum M_{BCW}$ = The sum of clockwise moments about <em>B</em>

$\sum M_{BCCW}$ = The sum of counterclockwise moments about <em>B</em>

Therefore, we have;

$\sum M_{BCW}$ = 2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81

$\sum M_{BCCW}$ = $F_R$ × √(6² - 2²)

Therefore, we get;

2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81 = $F_R$ × √(6² - 2²)

$F_R$ = (2 × (2/6) × 10.0 × 9.81 + 3.0 × (2/6) × 70 × 9.81)/(√(6² - 2²)) ≈ 132.95

The reaction force on the wall, $F_R$ ≈ 132.95 N

We note that the magnitude of the reaction force at the roof, $F_R$ = The magnitude of the frictional force of bottom of the ladder on the floor, $F_f$ but opposite in direction

Therefore;

$F_R$ = $-F_f$

$F_f$ = - $F_R$ ≈ -132.95 N

Similarly, at equilibrium, we have;

∑Fₓ = $\sum F_y$ = 0

The vertical component of the forces acting on the ladder are, (taking forces acting upward as positive;

$\sum F_y$ = -70.0 × 9.81 - 10 × 9.81 + $F_{By}$

∴ The upward force acting at the bottom, $F_{By}$ = 784.8 N

Therefore;

The magnitudes of the forces at the ladder top and bottom are;

At the top;

$\mathbf{F_{Top, \ x}}$ = $F_R$ ≈ 132.95 N←

$\mathbf{F_{Top, \ y}}$ = 0 (The surface upon which the ladder rest at the top is frictionless)

At the bottom;

$\mathbf{F_{Bottom, \ x}}$ = $F_f$ ≈ -132.95 N →

$\mathbf{F_{Bottom, \ y}}$ = $F_{By}$ = 784.8 N ↑

Learn more about equilibrium of forces here;

brainly.com/question/16051313

8 0

3 years ago