If object a has more mass than object b does object a contain more matter?

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

DENIUS [597]

Explanation:

It is given that,

The acceleration of the toboggan, $a=1.8\ m/s^2$

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

At t = 1 s

$s=\dfrac{1}{2}\times 1.8\times 1^2$

$s_1=0.9\ m$

At t = 2 s

$s=\dfrac{1}{2}\times 1.8\times 2^2$

$s_2=3.6\ m$

At t = 3 s

$s=\dfrac{1}{2}\times 1.8\times 3^2$

$s_3=8.1\ m$

Hence, this is the required solution.

4 0

3 years ago

Bruno the bat flies at a speed of 0.5 m/s in circle of radius 1 m. What is his acceleration?

beks73 [17]

Answer:

Acceleration is 0.25m/s^2

Explanation:

Given the following :

Speed = 0.5m/s

Radius(r) of circle = 1m

Acceleration round a circular path is given as :

a = v^2 / r

Where

a = acceleration of the body

v = speed / velocity

r = radius

Therefore,

a = v^2 / r

a = (0.5)^2 / 1

a = 0.25m/s^2

7 0

2 years ago

The masses are m1 = m, with initial velocity 2v0, and m2 = 7.4m, with initial velocity v0. Due to the collision, they stick toge

lesya [120]

Answer:

Loss, $\Delta E=-10.63\ J$

Explanation:

Given that,

Mass of particle 1, $m_1=m =0.66\ kg$

Mass of particle 2, $m_2=7.4m =4.884\ kg$

Speed of particle 1, $v_1=2v_o=2\times 6=12\ m/s$

Speed of particle 2, $v_2=v_o=6\ m/s$

To find,

The magnitude of the loss in kinetic energy after the collision.

Solve,

Two particles stick together in case of inelastic collision. Due to this, some of the kinetic energy gets lost.

Applying the conservation of momentum to find the speed of two particles after the collision.

$m_1v_1+m_2v_2=(m_1+m_2)V$

$V=\dfrac{m_1v_1+m_2v_2}{(m_1+m_2)}$

$V=\dfrac{0.66\times 12+4.884\times 6}{(0.66+4.884)}$

V = 6.71 m/s

Initial kinetic energy before the collision,

$K_i=\dfrac{1}{2}(m_1v_1^2+m_2v_2^2)$

$K_i=\dfrac{1}{2}(0.66\times 12^2+4.884\times 6^2)$

$K_i=135.43\ J$

Final kinetic energy after the collision,

$K_f=\dfrac{1}{2}(m_1+m_2)V^2$

$K_f=\dfrac{1}{2}(0.66+4.884)\times 6.71^2$

$K_f=124.80\ J$

Lost in kinetic energy,

$\Delta K=K_f-K_i$

$\Delta K=124.80-135.43$

$\Delta E=-10.63\ J$

Therefore, the magnitude of the loss in kinetic energy after the collision is 10.63 Joules.

7 0

2 years ago

An object has 90,000 J of kinetic energy and is moving at 12 m/s. What is the objects mass?

trapecia [35]

Answer:

plug it into your calculator

Explanation:

Kinetic energy = KE = Joules = J

KE= 1/2mv^2

90,000 = 1/2 m (12^2)

180,000 = m (144)

m = 180,000 / 144

4 0

3 years ago

A resistor, an inductor, and ideal battery, and a switch are connected in series. Just after the switch is closed, which of the

Marianna [84]

Answer:

b) the current and the voltage across the resistor

Explanation:

As soon as we close the switch then the current in the circuit will be zero because the inductor connected in series will not allow to change the current through it due to its inertial property

Since initial current through the inductor is zero so after connecting the key the current will still remains zero and then it will start increasing

The equation for current is given as

$i = i_o(1 - e^{-Rt/L})$

so as soon as the key is closed the current in the circuit is zero and hence the voltage across the resistor is also zero as we know

$V_r = i R = 0$

so correct answer will be

b) the current and the voltage across the resistor

5 0

3 years ago