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tiny-mole [99]
1 year ago
7

A 2-kg toy car accelerates from 0 to 5 m/s2. It

Physics
1 answer:
yan [13]1 year ago
8 0
10 joules of work is done by the object
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A frog leaps up from the ground and lands on a step 0.1 m above the ground 2 s later. We want to find the vertical velocity of t
zhuklara [117]

Answer:

<em>You would use the kinematic formula:</em>

    \Delta y=V_{0y}\times t-g\times t^2/2

Explanation:

The upwards vertical motion is ruled by the equation:

        y=y_0+V_{0y}\times t-g\times t^2/2

Where:

       y \text{ is the position at the time }t:y=0.1m

       y_0\text{ is the initial position: }y_0=0

       t=2s

       g\text{ is the gravitational acceleration: }\approx 9.8m/s^2

       V_{0y}\text{ is the initial vertical velocity}

Naming Δy = y - y₀, the equation becomes:

      \Delta y=V_{0y}\times t-g\times t^2/2

Then, you just need to substitute with Δy = 0.1m, t = 2s, and g = 9.8m/s², ans solve for the intital vertical velocity.

7 0
2 years ago
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Shutting the fluid discharge of an air-operated reciprocating pump will cause the pump to ?
Misha Larkins [42]
Had to look for the options and here is my answer. What happens when the fluid discharge of an air-operated reciprocating pump is shut, this will cause the pump to OVERSTROKE. Overstroke happens when the engine is switching in a normally-closed manner.  
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How does giving this award help the BLM more
Kazeer [188]

Answer:

Number 3

Explanation:

6 0
2 years ago
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Calculate the change in the kinetic energy (KE) of the bottle when the mass is increased. Use the formula KE = mv2, where m is t
Aliun [14]

kinetic energy is given as

KE = (0.5) m v²

given that : v = speed of the bottle in each case =  4 m/s

when m = 0.125 kg

KE = (0.5) m v² =  (0.5) (0.125) (4)² = 1 J

when m = 0.250 kg

KE = (0.5) m v² =  (0.5) (0.250) (4)² = 2 J

when m = 0.375 kg

KE = (0.5) m v² =  (0.5) (0.375) (4)² = 3 J

when m = 0.0.500 kg

KE = (0.5) m v² =  (0.5) (0.500) (4)² = 4 J

6 0
1 year ago
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When a car is 100 meters from its starting position traveling at 60.0 m/s., it starts braking and comes to a stop 350 meters fro
NISA [10]
Remember your kinematic equations for constant acceleration. One of the equations is x_{f} =  x_{i} +  v_{i}(t) + \frac{1}{2} at^{2}, where x_{f} = final position, x_{i} = initial position, v_{i} = initial velocity, t = time, and a = acceleration. 

Your initial position is where you initially were before you braked. That means x_{i} = 100m. You final position is where you ended up after t seconds passed, so x_{f} = 350m. The time it took you to go from 100m to 350m was t = 8.3s. You initial velocity at the initial position before you braked was v_{i} = 60.0 m/s. Knowing these values, plug them into the equation and solve for a, your acceleration:
350\:m = 100\:m + (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\&#10;250\:m = (60.0\:m/s)(8.3\:s) + \frac{1}{2} a(8.3\:s)^{2}\\&#10;250\:m = 498\:m +34.445\:s^{2}(a)\\&#10;-248\:m = 34.445\:s^{2}(a)\\&#10;a \approx -7.2 \: m/s^{2}

Your acceleration is approximately -7.2 \: m/s^{2}.
4 0
2 years ago
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