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3 years ago

3. Calculate the force it would take to accelerate a 50 ka bike at a rate of 3 m/s2 (6 points)

Physics

1 answer:

qwelly [4]3 years ago

8 0

Answer:

150 N

Explanation:

Given that,

Acceleration (a) = 3 m/s²
Mass of the bike (m) = 50 kg

We are asked to calculate force required.

$\longrightarrow$ F = ma

$\longrightarrow$ F = (50 × 3) N

$\longrightarrow$ F = 150 N

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Three tiny charged metal balls are arranged on a straight line. The middle ball is positively charged and the two outside balls

Dmitrij [34]

Answer:

$(a) 189.23 N$ , $(b) 47.31 N$ and $(c) 141.92 N$ .

Explanation:

Three balls are shown in figure having charge $q=1.45 \mu C$ . The middle ball, $B$ , is positively charged having charge $+q$ , and the remaining two outside balls, $A$ and $C$ , are negatively charged having charged $-q$ as shown.

$AC=20 cm$ and $AB=BC=10$ cm as B is the mid-point of AC.

Let $d_1=AC=20\times 10^{-3}m$ and $d_2=AB=BC=10\times 10^{-3}m$

From Coulomb's law, the magnitude of the force, $F$ , between two point charges having magnitudes $q_1 \& q_2$ , separated by distance, $d$ , is

$F=\frac {1}{4\pi\epsilon_0}\frac {q_1q_2}{d^2}\;\cdots (i)$

where, $\epsilon_0$ is the permittivity of free space and

$\frac {1}{4\pi\epsilon_0}=9\times 10^9$ in SI units.

This force is repulsive for the same nature of charges and attractive for the different nature of charges.

Now, Using equations(i),

(a) The magnitude of attraction force between balls A and B is

$F_{AB}=F_{BC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_2)^2}$

$\Rightarrow F_{AB}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(10\times 10^{-3}\right)^2}$

$\Rightarrow F_{AB}=189.23 N$

(a) The magnitude of the repulsive force between balls A and C is

$F_{AC}= \frac {1}{4\pi\epsilon_0}\frac {qq}{(d_1)^2}$

$\Rightarrow F_{AC}= 9\times 10^9}\frac {1.45\times 10^{-6}\times1.45\times 10^{-6}}{\left(20\times 10^{-3}\right)^2}$

$\Rightarrow F_{AC}=47.31 N$

$F_{net}=189.23-47.31 N$

$\Rightarrow F_{net}=141.92 N$

4 0

3 years ago

A permanent magnet creates a magnetic field at the origin with strength Bperm-1T. A current-carrying wire is oriented such that

bazaltina [42]

Answer:

Part a)

$B_{net} = (1 + 4 \times 10^{-6})T$

Part b)

$B_{net} = (1 - 4 \times 10^{-6})T$

Explanation:

Part a)

Since the two magnetic field is in same direction

so the net magnetic field is algebraic sum of magnetic field due to both

so here magnetic field of wire is given as

$B = \frac{\mu_0 i}{2\pi r}$

here we know that

I = 2 A

r = 5 cm

so we will have

$B = \frac{2 \times 10^{-7} (2)}{0.05}$

$B = 4 \times 10^{-6} T$

So net magnetic field is given as

$B_{net} = (1 + 4 \times 10^{-6})T$

Part b)

When direction of current is reversed then the direction of magnetic field is also reversed

So we will have

$B_{net} = (1 - 4 \times 10^{-6})T$

8 0

3 years ago

Help please!!!!!!!!!!!!

rodikova [14]

Burning wood would be the answer

7 0

3 years ago

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Which lobes of the brain receive the input that enables you to feel someone scratching your back? (2 points)

andrew-mc [135]

Answer: Parietal

Explanation: The parietal lobe is where the primary somatosensory cortex is located. This cortex is where all tactile stimulation is processed in the brain and allows to you detect/feel someone scratching your back.

3 0

3 years ago

Bumper car A (281 kg) moving

emmainna [20.7K]

Answer:

1.88 m/s

Explanation:

from elastic collision

KE before collision = KE after collision

8 0

3 years ago

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