All categories

2 years ago

Sean pushes a box with a force of 100 N. Christian pushes an identical box with a force of

Physics

1 answer:

kap26 [50]2 years ago

5 0

Answer:

Christian

Explanation:

Given that

Force applied by Sean ,F₁=100 N

Force applied by Christian ,F₂=200 N

Lets take mass of the box = m kg

As we know that from second law of Newton's

F= m a

F=Force,m=mass ,a=acceleration

$F_1=ma_1$

$100=m\times a_1$

$a_1=\dfrac{100}{m}\ m/s^2$

$F_2=ma_2$

$200=m\times a_2$

$a_2=\dfrac{200}{m}\ m/s^2$

From the above we can say that acceleration a₂ is greater than a₁ is for the same mass of the box.

Therefore Christian will acceleration more for the box.

You might be interested in

Qual a capacidade térmica de um objeto que, ao receber 10000 cal de energia, tem sua temperatura elevada de 25°C para 75°C?

Stolb23 [73]

Answer:

$200 cal/^{\circ}C$

Explanation:

When heat energy is supplied to an object, the temperature of the object increases according to the equation:

$Q=C\Delta T$

where

Q is the heat supplied

C is the heat capacity of the object

$\Delta T$ is the change in temperature

In this problem we have:

$Q=10,000 cal$ is the energy supplied

$\Delta T=75C-25C=50C$ is the change in temperature of the object

Therefore, the heat capacity of the object is:

$C=\frac{Q}{\Delta T}=\frac{10,000}{50}=200 cal/^{\circ}C$

6 0

2 years ago

A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud

IceJOKER [234]

Answer: 1.51 km

Explanation:

Coulomb's Law: The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or, $\vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}$

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

Given:

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

So,

$\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514 \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}$

Therefore, the separation between the two charges (r) = 1.51 km

3 0

2 years ago

A ball is released from the top of a hill. How fast is the ball going when it reaches the base of the hill? Approximate g as 10

irina [24]

First I’ll show you this standard derivation using conservation of energy:
Pi=Kf,
mgh = 1/2 m v^2,
V = sqrt(2gh)
P is initial potential energy, K is final kinetic, m is mass of object, h is height from stopping point, v is final velocity.
In this case the height difference for the hill is 2-0.5=1.5 m. Thus the ball is moving at sqrt(2(10)(1.5))=
5.477 m/s.

5 0

2 years ago

Read 2 more answers

What is the velocity of the particle when its acceleration is zero?

Zigmanuir [339]

Its velocity would be constant

3 0

3 years ago

Gary saved dimes and nickels at a ratio of 5:7. Then, he saved 20% more dimes and 3 times the original number of nickels. If he

TiliK225 [7]

Answer:

i think 53 or 56